2
$\begingroup$

We call a number algebraic if and only if it is the solution of a polynomial with integer coefficients. A number (complex or real) is transcendental if and only if it is not algebraic.

A while back while reading about transcendental numbers (and the open problem of whether Catalan's constant is transcendental) I recall reading about some strange number, expressed as a series, which actually turned out to be algebraic (with the polynomial being quite long and crazy).

Can anyone give me a reference to it (or perhaps to another similar algebraic number)? I am willing to consider any algebraic number, which was originally defined as a 'nice' series, has no trivial expression as a quadratic surd (or sum of such), and has a long and crazy polynomial, as a satisfactory answer.

Edit: I think this (the proof that the number was algebraic) was a rather new result, i.e. in the past 10 (or at most 20) years.

$\endgroup$
2
$\begingroup$

Take any root of any polynomial "quite long and crazy". You can approximate the root using a sequence of rational numbers and then easily find a series with sum = the root.

Nice example: $$\sqrt 2 = \sum_{k=0}^\infty(-1)^{k+1}\frac{(2k-3)!!}{(2k)!!}$$ (write the Taylor series of $\sqrt{1+x}$)

$\endgroup$
  • $\begingroup$ Thank you for the example. This is not what I had in mind, as the number was not expressible as a quadratic surd (else it would not have been an open problem whether or not it was algebraic for so long!), but if no-one gives a better example in a couple of days, I'll accept yours. $\endgroup$ – Simon_Peterson Jan 19 '16 at 18:48
  • $\begingroup$ @Simon_Peterson, you can generalize the example to any $n$-th root. Sum two (or more) roots: $\root{127}\of 2+\root{999}\of 2$ is an algebraic number with a long and crazy minimal polynomial that can be expressed with a nice series. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 19 '16 at 18:55
  • $\begingroup$ Yes, I know, but the point was that the number I am looking for was originally defined as a series, and it was unknown for a long time whether it was algebraic or not. $\endgroup$ – Simon_Peterson Jan 19 '16 at 18:59
  • $\begingroup$ @Simon_Peterson, in any case, my point is that the property "algebraic and sum of a series" is almost trivial. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 19 '16 at 18:59
  • $\begingroup$ @Simon_Peterson, OK, now I see your point. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 19 '16 at 19:01

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.