Pick two random numbers from the interval ; independent, uniformly distributed. Normalize them to have mean zero, which simply means subtracting their mean from each. Repeat many times. Plot the histogram of all numbers obtained in the process.
No surprise here. In effect this is the distribution of with independent and uniformly distributed over . The probability density function of is found via convolution, and the convolution of with itself is a triangular function.
Repeat the same with four numbers , again subtracting the mean. Now the distribution looks vaguely bell-shaped.
With ten numbers or more, the distribution is not so bell-shaped anymore: the top is too flat.
The mean now follows an approximately normal distribution, but the fact that it’s subtracted from uniformly distributed amounts to convolving the Gaussian with . Hence the flattened top.
What if we use the median instead of the mean? With two numbers there is no difference: the median is the same as the mean. With four there is.
That’s an odd-looking distribution, with convex curves on both sides of a pointy maximum. And with points it becomes even more strange.
k = 10
A = rand(200000,k)
A = A - median(A,'c').*.ones(1,k)
The closed disk of radius has area . But it happens to contain only points with integer coordinates. Here is a picture of one quarter of this disk.
The radius is somewhat notable is that the discrepancy of between the area and the number of integer points is unusually large. Here is the plot of the absolute value of the difference |area-points| as a function of integer radius . The curve in red is , which is an experimentally found upper bound for the discrepancy in this range of .
On the scale up to , the upper bound is , and the radii bumping against this bound are and . The exponent begins to resemble the conjectural in the Gauss circle problem.
Finally, over the interval the upper bound comes out as . The exponent looks good.
This little numerical experiment in Matlab involved using the convex hull function convhull on log-log scale. The function identifies the vertices of the convex hull, which is a polygon. I pick the side of the polygon lying over the midpoint of the range; this yields a linear upper bound for the set of points. On the normal scale, this line becomes a power function. Matlab code is given below; it’s divided into three logical steps.
Find the difference between area and the number of lattice points
a = 1000;
b = 3000;
R = a:b;
E = ;
for n = a:b
[X,Y] = meshgrid(1:n, 1:n);
pts = 4*n + 1 + 4*nnz(X.^2+Y.^2<=n^2);
E = [E, max(1,abs(pts - pi*n^2))];
Pick a suitable side of log-log convex hull
ix = convhull(log(R), log(E));
k = numel(ix);
k = k-1;
Plot the result and output the parameters of the upper bound
R1 = R(ix(k)); E1 = E(ix(k));
R2 = R(ix(k+1)); E2 = E(ix(k+1));
b = log(E1/E2)/log(R1/R2);
a = E1/R1^b;
plot(R, E, '.');
plot(R, a*R.^b , 'r-');
fprintf('a = %.3f, b = %.3f\n', a, b);
If one picks two real numbers from the interval (independent, uniformly distributed), their sum has the triangular distribution.
The sum of three such numbers has a differentiable probability density function:
And the density of is smoother still: the p.d.f. has two
As the number of summands increases, these distributions converge to normal if they are translated and scaled properly. But I am not going to do that. Let’s keep the number of summands to four at most.
The p.d.f. of is a piecewise polynomial of degree . Indeed, for the density is piecewise constant, and the formula
provides the inductive step.
For each , the translated copies of function form a partition of unity:
The integral recurrence relation gives an easy proof of this:
And here is the picture for the quadratic case:
A partition of unity can be used to approximate functions by piecewise polynomials: just multiply each partition element by the value of the function at the center of the corresponding interval, and add the results.
Doing this with amounts to piecewise linear interpolation: the original function is in blue, the weighted sum of hat functions in red.
With we get a smooth curve.
Unlike interpolating splines, this curve does not attempt to pass through the given points exactly. However, it has several advantages over interpolating splines:
Is easier to calculate; no linear system to solve;
Let be some norm on . The norm induces a metric, and the metric yields a notion of curve length: the supremum of sums of distances over partitions. The unit circle is a closed curve; how small can its length be under the norm?
For the Euclidean norm, the length of unit circle is . But it can be less than that: if is a regular hexagon, its length is exactly . Indeed, each of the sides of is a unit vector with respect to the norm defined by , being a parallel translate of a vector connecting the center to a vertex.
To show that cannot be beaten, suppose that is the unit circle for some norm. Fix a point . Draw the circle ; it will cross at some point . The points are vertices of a hexagon inscribed in . Since every side of the hexagon has length , the length of is at least .
It takes more effort to prove that the regular hexagon and its affine images, are the only unit circles of length ; a proof can be found in Geometry of Spheres in Normed Spaces by Juan Jorge Schäffer.
If a function is continuous, then its graph is closed. The converse is false: a counterexample is given by any extension of to the real line.
The Closed Graph Theorem of functional analysis states that a linear map between Banach spaces is continuous whenever its graph is closed. Although the literal extension to nonlinear maps fails, it’s worth noting that linear maps are either continuous or discontinuous everywhere. Hence, if one could show that a nonlinear map with a closed graph has at least one point of continuity, this would be a nonlinear version of the Closed Graph Theorem.
Here is an example of a function with a closed graph and an uncountable set of discontinuities. Let be a closed set with empty interior, and define
For a general function, the set of discontinuities is an set. When the graph is closed, we can say more: the set of discontinuities is closed. Indeed, suppose that a function is bounded in a neighborhood of but is not continuous at . Then there are two sequences and such that both sequences and converge but have different limits. Since at least one of these limits must be different from , the graph of is not closed. Conclusion: a function with a closed graph is continuous at if and only if it is bounded in a neighborhood of . In particular, the set of discontinuities is closed.
Furthermore, the set of discontinuities has empty interior. Indeed, suppose that is discontinuous at every point of a nontrivial closed interval . Let ; this is a closed bounded set, hence compact. Its projection onto the -axis is also compact, and this projection is exactly the set . Thus, is closed. The set has empty interior, since otherwise would be continuous at its interior points. Finally, , contradicting the Baire Category theorem.
Summary: for closed-graph functions on , the sets of discontinuity are precisely the closed sets with empty interior. In particular, every such function has a point of continuity. The proof works just as well for maps from to any metric space.
However, the above result does not extend to the setting of Banach spaces. Here is an example of a map on a Banach space such that whenever ; this property implies that the graph is closed, despite being discontinuous everywhere.
Let the space of all bounded functions with the supremum norm. Let be an enumeration of all rational numbers. Define the function separately on each subinterval , as
For any two distinct elements of there is a point and a number such that is strictly between and . According to the definition of this implies that the functions and take on different values at the point . Thus the norm of their difference is .
So much for Nonlinear Closed Graph Theorem. However, the space in the above example is nonseparable. Is there an nowhere continuous map between separable Banach spaces such that its graph is closed?
Adding 2015 Fellows changed a few things compared to 2014 rankings. UCLA and Rutgers rise from 2nd and 3rd to tie Berkeley for the first place. Syracuse loses nine places, dropping from #334(tie) to #343(tie).
1. Rutgers The State University of New Jersey New Brunswick: 34
In fact, it worked much better than I expected. Not only if of constant sign, but so are and . Indeed,
is always negative,
is always positive,
is always negative. The sign becomes less obvious with the fourth derivative,
because the triangle inequality isn’t conclusive now. But the amplitude of is , and .
So, it seems that is completely monotone, meaning that for all and for all . But we already saw that this sign pattern can break after many steps. So let’s check carefully.
Direct calculation yields the neat identity
With its help, the process of differentiating the function can be encoded as follows: , , then and . The presence of is disconcerting because the harmonic series diverges. But orthogonality helps: the added vector is orthogonal to .
The above example, rewritten as , corresponds to starting with . I calculated and plotted iterations: the points are joined by piecewise linear curve.
The total length of this curve is infinite, since the harmonic series diverges. The question is, does it stay within the unit disk? Let’s find out. By the above recursion,
Hence, the squared magnitude of will always be less than
with being . The infinite product evaluates to (explained here), and thus the polygonal spiral stays within the disk of radius . In conclusion,
where the trigonometric function has amplitude strictly less than . Since the expression on the right is positive, is completely monotone.
The plot was generated in Sage using the code below.
a,b,c,d = var('a b c d')
a = 0
b = 1/2
l = [(a,b)]
for k in range(1,10000):
c = a-b/k
d = b+a/k
a = c
b = d
The series diverges. A common way to make it convergent is to replace each with a power of ; the new series will converge when and maybe its sum will have a limit as . And indeed,
which tends to as approaches from the left.
Things get more interesting if instead of consecutive integers as exponents, we use consecutive powers of :
On most of the interval it behaves just like the previous one:
But there appears to be a little blip near . Let’s zoom in:
And zoom in more:
This function was considered by Hardy in 1907 paper Theorems concerning infinite series. On pages 92–93 he shows that it “oscillates between finite limits of indetermination for “. There is also a footnote: “The simple proof given above was shown to be by Mr. J. H. Maclagan-Wedderburn. I had originally obtained the result by means of a contour integral.”
Okay, but what are these “finite limits of indetermination”? The Alternating Series Estimation shows for , but the above plots suggest that oscillates between much tighter bounds. Let’s call them and .
Since , it follows that as . Hence, . In other words, and are symmetric about . But what are they?
I don’t have an answer, but here is a simple estimate. Let and observe that
The function is not hard to understand: its graph is a parabola.
Since is positive on , any of the terms in the sum (1) gives a lower bound for . Each individual term is useless for this purpose, since it vanishes at . But we can pick in depending on .
Let be the unique solution of the equation . It could be written down explicitly, but this is not a pleasant experience; numerically . For every there is an integer such that , namely the smallest integer such that . Hence,
which gives a nontrivial lower bound and symmetrically . Frustratingly, this falls just short of neat and .
One can do better than (2) by using more terms of the series (1). For example, study the polynomial and find a suitable interval on which its minimum is large (such an interval will no longer be symmetric). Or use consecutive terms of the series… which quickly gets boring. This approach gives arbitrarily close approximations to and , but does not tell us what these values really are.
To generate a random number uniformly distributed on the interval , one can keep tossing a fair coin, record the outcomes as an infinite sequence of 0s and 1s, and let . Here is a histogram of samples from the uniform distribution… nothing to see here, except maybe an incidental interference pattern.
Let’s note that where has the same distribution as itself, and is independent of . This has an implication for the (constant) probability density function of :
because is the p.d.f. of and is the p.d.f. of . Simply put, is equal to the convolution of the rescaled function with the discrete measure .
Let’s iterate the above construction by letting each be uniformly distributed on instead of being constrained to the endpoints. This is like tossing a “continuous fair coin”. Here is a histogram of samples of ; predictably, with more averaging the numbers gravitate toward the middle.
This is not a normal distribution; the top is too flat. The plot was made with this Scilab code, putting n samples put into b buckets:
n = 1e6
b = 200
z = zeros(1,n)
for i = 1:10
z = z + rand(1,n)/2^i
c = histplot(b,z)
If this plot too jagged, look at the cumulative distribution function instead:
It took just more line of code: plot(linspace(0,1,b),cumsum(c)/sum(c))
Compare the two plots: the c.d.f. looks very similar to the left half of the p.d.f. It turns out, they are identical up to scaling.
Let’s see what is going on here. As before, where has the same distribution as itself, and the summands are independent. But now that is uniform, the implication for the p.d.f of is different:
This is a direct relation between and its antiderivative. Incidentally, if shows that is infinitely differentiable because the right hand side always has one more derivative than the left hand side.
To state the self-similarity property of in the cleanest way possible, one introduces the cumulative distribution function (the Fabius function) and extends it beyond by alternating even and odd reflections across the right endpoint. The resulting function satisfies the delay-differential equation : the derivative is a rescaled copy of the function itself.
Since vanishes at the even integers, it follows that at every dyadic rational, all but finitely many derivatives of are zero. The Taylor expansion at such points is a polynomial, while itself is not. Thus, is nowhere analytic despite being everywhere .
This was, in fact, the motivation for J. Fabius to introduce this construction in 1966 paper Probabilistic Example of a Nowhere Analytic -Function.