Steven R. Dunbar
Department of Mathematics
203 Avery Hall
University of Nebraska-Lincoln
Lincoln, NE 68588-0130
Probability Theory and Stochastic Processes
Steven R. Dunbar
Mathematically Mature: may contain mathematics beyond calculus with proofs.
Can you think of a sequence or a process that approximates ? What is the intuition or reasoning behind that sequence?
Wallis’ Formula is the amazing limit
Another way to write this is
A “closed form” expression for the product in Wallis formula is
Note that Wallis Formula is equivalent to saying that the “central binomial term” has the asymptotic expression
See the subsection Central Binomial below for a proof of an equivalent inequality.
In the form
it is easy to see that the sequence is increasing since . Figure 1 illustrates the increasing sequence.
Doing a numerical linear regression of versus on the domain to indicates that approaches at a rate which is .
Proof. Consider . Integrating by parts with and shows
Gathering terms, we get .
Now so recursively , and inductively
Likewise , and , and inductively
For , , so , implying in turn that . Then
The following proof is adapted and expanded from . The proof uses only basic algebra, the Pythagorean Theorem, and the formula for the area of a circle of radius . Another important property used implicitly is the completeness property of the reals.
Deﬁne a sequence of numbers by and for ,
These are the reciprocals of the subsequence deﬁned in equation (4).
The partial products of Wallis’ formula (1) with an odd number of terms in the numerator are
while those with an even number of factors in the numerator are of the form
Interpret as an empty product. Since , clearly . Since , clearly . Also, . Therefore
for any . Furthermore, for ,
from which it follows that
For convenience, deﬁne so that inequality (8) holds also for . Denote the successive diﬀerence so and for
Proof. Make the substitutions
the right side of the proposed identity (9) becomes
Collecting terms, this simpliﬁes to . □
Now divide the positive quadrant of the -plane into rectangles by drawing the vertical lines and the horizontal lines for all . Let be the rectangle with lower left corner and upper right corner . The area of is . Thus the identity states that the total area of the rectangles for which is . Let be the polygonal region consisting of all rectangles for which . Hence the area of is .
The outer corners of are the points for which and . By the Pythagorean theorem, the distance of such a point to the origin is . By (8)
bounds this distance from above. Similarly the inner corners of are the points for which and . The distance of such a point to the origin is bounded from below by
Therefore, contains a quarter circle of radius and is contained in a quarter circle of radius . See Figure 2 for a diagram of the polygonal region and the corresponding inner and outer quarter circles for .
The area of a quarter circle of radius is and the area of is . This leads to the bounds
from which it follows that
Then as , and both approach .
This subsection gives a detailed proof that Wallis’ Formula gives an explicit inequality bound on the central binomial term that in turn implies the asymptotic formula for the central binomial coeﬃcient. The derivation in . motivates this proof.
Start from the expansion (2) for the Central Binomial Coeﬃcient:
Rearrange it to
and use the deﬁnitions (6) and (7) of the partial products of the Wallis formula to obtain
Rearrange the inequality (10) to obtain
and taking square roots and slightly rearranging again
To simplify, take a series expansion of the square roots of the rational expressions and truncate, leaving
Proof. See [7, page 312] for the proof. The article in the Mathworld.com article on http://mathworld.wolfram.com/Sine.html. also gives references to Edwards 2001, pages 18 and 47; and Borwein et al. 2004, page 5. □
Although the common proof uses complex analysis, as in the texts cited above, a proof using only elementary analysis is possible. The following proof is adapted from .
Proof. Start with the deﬁnition of the Chebyshev polynomials from the trigonometric identity . Then
Together with the sine product identity
this inductively shows that there is a polynomial of degree such that
(In fact, using the deﬁnition for the Chebyshev polynomials of the second kind, it is easy to show that .) Substituting and noting that shows that has zeros at for . These zeros are distinct, so has no other zeros, then
and changing variables
The goal now is to estimate the product terms. For all real , and therefore . For , the choice gives . Then for every collection of numbers , we have
If also , then we also know that
Let and be positive integers with . Take such that and . Then deﬁne by
Use (11) by dividing the leading factor and the ﬁrst factors onto the left side to obtain
Then use the ﬁrst, third and ﬁfth terms of (12) to see that
For we have , so we see that
Thus it follows from (15) that
Let so that
Now let and we obtain for
For this equality is also true. □
The following somewhat probabilistic proof of Euler’s inﬁnite product formula is adapted from .
Proof. Start with the integral
and integrate by parts once:
Integrate by parts again:
After integration by parts times:
Then by dominated convergence it follows that
Note that limit deﬁnition of the Gamma function is also http://dlmf.nist.gov/5.8 E1.. It follows that for that
For and independent exponential random variables, both with expectation we have that
and likewise . Then using the independence and symmetry
The distribution function for the random variable is
for . Then the probability density is
This gives by integration by parts, for , that
The last integral is . For example, this is formula 613, page 445 in the 18th Edition of the CRC Standard Mathematical Tables. It can also be done with contour integration in the complex plane
A standard integration-by-parts gives the fundamental Gamma function recursion . Using this to deﬁne for , it follows that for all real ,
Combining all results above,
Proof. Substitute in the product expansion of . □
Remark. In  Ciaurri uses Tannery’s Theorem for Inﬁnite Products, a trig identity for cotangent and tangent, and Wallis formula to provide an elementary proof of product expansion of the sine function. In this sense, the product expansion of the sine function is equivalent to Wallis’ formula.
This section gives a geometric interpretation of Wallis’ formula, constructing a subset of the unit square with area . The construction is reminiscent of the construction of the Sierpinski Triangle.
Take a unit square and remove a square of area from the center. Label the result as with area . Next remove a square of area from the center of each of the small sub-squares of area constituting and call the result with area . Continuing by removing squares of area from each sub-square of area surrounding the hole removed at stage , construct region with area . Continue to create , and so on. Call the limiting region which has area .
The side length of the removed squares at stage is . After stages, squares have been removed, and the stage holes have side length . It will be impractical to illustrate more than stages of this iterative process.
Here area means the Lebesgue measure of the limiting region . However does not contain any product set with , each with positive Lebesgue measure. To see this, consider a maximal product subset of , for example (or equivalently although from here on, consider only the product sets with ﬁrst factor ). This product subset has measure . A similar maximal product subset of has measure . Continuing, a maximal product subset of has measure .
Thus, is an example of a subset of the plane with positive Lebesgue measure, but not admitting any product subset with positive Lebesgue measure. Note also that is a “square-like” region with area equal to a circle with radius . In that sense, this is a solution to the ancient problem of “squaring the circle”.
This section is adapted from a sketch of the proof of Wallis’ Formula in Kazarinoﬀ  and the short note by Wästlund . The elementary proofs of the sine product are from  and . The geometric interpretation of Wallis’ Formula is adapted from .
approaches by calculating values and using regression.
Explicitly ﬁnd and plot them on .
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Steve Dunbar’s Home Page, http://www.math.unl.edu/~sdunbar1
Email to Steve Dunbar, sdunbar1 at unl dot edu
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