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September 21, 2018

A Pattern That Eventually Fails

Posted by John Baez

Sometimes you check just a few examples and decide something is always true. But sometimes even 1.5×10 431.5 \times 10^{43} examples is not enough.

You can show that

0 sinttdt=π2\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, d t = \frac{\pi}{2} }

0 sinttsin(t101)t101dt=π2\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, d t = \frac{\pi}{2} }

0 sinttsin(t101)t101sin(t201)t201dt=π2\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, \frac{\sin \left(\frac{t}{201}\right)}{\frac{t}{201}} \, d t = \frac{\pi}{2} }

0 sinttsin(t101)t101sin(t201)t201sin(t301)t301dt=π2\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, \frac{\sin \left(\frac{t}{201}\right)}{\frac{t}{201}} \, \frac{\sin \left(\frac{t}{301}\right)}{\frac{t}{301}} \, d t = \frac{\pi}{2} }

and so on.

It’s a nice pattern. But it doesn’t go on forever! In fact, Greg Egan showed the identity

0 sinttsin(t101)t101sin(t201)t201sin(t100n+1)t100n+1dt=π2\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, \frac{\sin \left(\frac{t}{201}\right)}{\frac{t}{201}} \cdots \, \frac{\sin \left(\frac{t}{100 n +1}\right)}{\frac{t}{100 n + 1}} \, d t = \frac{\pi}{2} }

holds when

n<15,341,178,777,673,149,429,167,740,440,969,249,338,310,889 n &lt; 15,341,178,777,673,149,429,167,740,440,969,249,338,310,889

but fails for all

n15,341,178,777,673,149,429,167,740,440,969,249,338,310,889. n \ge 15,341,178,777,673,149,429,167,740,440,969,249,338,310,889 .

It’s not as hard to understand as it might seem; it’s a special case of the infamous ‘Borwein integrals’. The key underlying facts are:

  • The Fourier transform turns multiplication into convolution.

  • The Fourier transform of sin(cx)/(cx)\sin(c x)/(c x) is a step function supported on the interval [c,c][-c,c].

  • The sum k=1 n1100k+1\displaystyle{\sum_{k = 1}^n \frac{1}{100k + 1}} first exceeds 11 when

n=15,341,178,777,673,149,429,167,740,440,969,249,338,310,889. n = 15,341,178,777,673,149,429,167,740,440,969,249,338,310,889.

For Greg’s more detailed explanation, based on that of Hanspeter Schmid, and for another famous example of a pattern that eventually fails, go here:

Posted at September 21, 2018 5:05 PM UTC

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6 Comments & 0 Trackbacks

Re: A Pattern That Eventually Fails

Very cool! This reminds me of the strong law of small numbers, except that I don’t usually think of 1.5×10 431.5\times 10^{43} as being very small. Of course, any finite number is small relative to infinity…

Posted by: Mike Shulman on September 21, 2018 6:11 PM | Permalink | Reply to this

Re: A Pattern That Eventually Fails

Don’t you want 100k+1 in the denominator of the 3rd bullet item?

Posted by: Silvio Levy on September 22, 2018 8:47 PM | Permalink | Reply to this

Re: A Pattern That Eventually Fails

Yes, I’ll fix that. Thanks!

Posted by: John Baez on September 22, 2018 9:36 PM | Permalink | Reply to this

Re: A Pattern That Eventually Fails

Perhaps another example is the Riemann Hypothesis.

Posted by: Jeffery Winkler on September 26, 2018 7:25 PM | Permalink | Reply to this

Re: A Pattern That Eventually Fails


In 2000, Gourdon and Demichel showed that the first 10,000,000,000,000 nontrivial zeros of the Riemann zeta function lie on the line Re(z) = 1/2, just as the Riemann hypothesis claims. They also checked two billion much larger zeros.

Normally this would seem pretty convincing. But some argue otherwise! According to Wikipedia:

At first, the numerical verification that many zeros lie on the line seems strong evidence for it. However, analytic number theory has had many conjectures supported by large amounts of numerical evidence that turn out to be false. See Skewes number for a notorious example, where the first exception to a plausible conjecture related to the Riemann hypothesis probably occurs around 10316; a counterexample to the Riemann hypothesis with imaginary part this size would be far beyond anything that can currently be computed using a direct approach. The problem is that the behavior is often influenced by very slowly increasing functions such as log log T, that tend to infinity, but do so so slowly that this cannot be detected by computation. Such functions occur in the theory of the zeta function controlling the behavior of its zeros; for example the function S(T) above has average size around (log log T)1/2. As S(T) jumps by at least 2 at any counterexample to the Riemann hypothesis, one might expect any counterexamples to the Riemann hypothesis to start appearing only when S(T) becomes large. It is never much more than 3 as far as it has been calculated, but is known to be unbounded, suggesting that calculations may not have yet reached the region of typical behavior of the zeta function.

Emphasis mine.

Posted by: John Baez on September 26, 2018 9:08 PM | Permalink | Reply to this

Re: A Pattern That Eventually Fails

And another example: a computer search has shown that among numbers less than a trillion, most common distance between successive primes is not 2 but 6. This goes on for quite a while longer…

… but Andrew Odlyzko, Michael Rubinstein and Marek Wolf have persuaded most experts that somewhere around x=1.742710 35x = 1.7427 \cdot 10^{35}, the most common gap between consecutive primes less than xx switches from 6 to 30.

And then later something else happens. I wrote about this here:

Posted by: John Baez on September 26, 2018 9:12 PM | Permalink | Reply to this

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