2
$\begingroup$

After having been devoting some time for many years to experimental mathematics, I am thinking to publish the details of some of my most fruitful computing workflows for discovering identities. I would like to be sure however that such identities are not trivial before spending too much time in describing my methods.

Could you have a look below at one of my most recent continued fraction and tell me if you think that a paper describing such a workflow (combination of various algorithms working together for building identities and selecting the relevant ones) would be worth being published?

Since the continued fractions I submitted here before were involving polynomial or periodic patterns, I selected the following result, which seems to be a little more sophisticated.

The partial numerators here are described at https://oeis.org/A026741 as “a(n) = n if n odd, n/2 if n even”.

$$ 2^{\alpha+1} \,\alpha \int_{x=0}^{1} \displaystyle \frac{x^{2\alpha}} {\left(1+x^2\right)^{\alpha+1}} \,\textrm{d}x = \cfrac{1} {1+\cfrac{1} {\alpha+\cfrac{1} {1+\cfrac{3} {\alpha+\cfrac{2} {1+\cfrac{5} {\alpha+\cfrac{3} {1+\cfrac{7} {\alpha+\cdots}}}}}}}} $$

Special values are related to $\pi$ when $\alpha$ is an integer, and to $\sqrt{2}$ when $\alpha$ is an half-integer.

I hope I made no mistake when writing down the formula, but you can have a quick check at the numerical counterpart by using the following Pari-GP code; both functions k (the continued fraction) and f (the left hand side) should give the same result for any positive value of $\alpha$.

K(b, a)=
{
  my(p1,p2,q1,q2,p,q);
  p1=1.;p2=0.;q1=0.;q2=1.;
  for(n=1,length(b),
    p = a[n]*p2+b[n]*p1;
    q = a[n]*q2+b[n]*q1;
    p1=p2;p2=p;q1=q2;q2=q);
  p/q
};
k(a) = K(concat([1], vector(256, n, n\(2-n%2))), concat([1], vector(256, n, if(n%2==1, a, 1))));
f(a) = 2^(a+1)*a*intnum(x = 0,1, x^(2*a)/(1+x^2)^(a+1));
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.