2
$\begingroup$

How do you/is it possible to express $a=\cfrac{1}{2+\cfrac{3}{4+\cfrac{5}{6+\cdots}}}$ in the form $\frac{p}{q}(k+\sqrt{n})$? I'm still in high school, so I'm not familiar with especially sophisticated approaches for evaluating infinitely continued fractions - I've been able to set up a recurrence relation for everything I've encountered so far, but after writing it out as $a(x)=\cfrac{x}{x+1+\cfrac{x+2}{x+3+\cdots}}$, I'm fairly sure that approach isn't going to work. I'm also not sure how to represent this in continued fraction notation, as each nested fraction has a unique numerator. It clearly converges to something, so can anyone point me in the right direction? $:)$

$\endgroup$
8
  • $\begingroup$ Take a look at this mathworld.wolfram.com/ContinuedFractionConstant.html . I believe some continued fractions aren't known to be expressed in simple forms. $\endgroup$
    – Jam
    Aug 11, 2014 at 13:54
  • 1
    $\begingroup$ The standard continued fraction of such a constant does not look to be (eventually) periodic like in the case of quadratic irrationals (Lagrange theorem, faculty.plattsburgh.edu/sam.northshield/09-0507.pdf) are you really sure the integers $p,q,k,n$ exist? $\endgroup$ Aug 11, 2014 at 14:00
  • 1
    $\begingroup$ For istance, I get $[0; 2, 1, 1, 1, 2, 1, 2, 10, 2, 2, 66,1,1,13,\ldots]$ as the standard continued fraction for $a$. $\endgroup$ Aug 11, 2014 at 14:02
  • 2
    $\begingroup$ It is definitely not quadratic since the continued fraction for a quadratic number is repeating $\endgroup$
    – Alice Ryhl
    Aug 11, 2014 at 15:10
  • 1
    $\begingroup$ @Darksonn Only the standard continued fraction need repeat. The standard continued fraction has its partial numerators all 1s. This does not rule out a nonrepeating, nonstandard continued fraction such as the one OP is asking about. $\endgroup$
    – MJD
    Aug 11, 2014 at 16:51

2 Answers 2

3
$\begingroup$

In view of the fact that the standard continued fraction expansion of $e$ is $$ e=2+\frac1{1+}\frac1{2+}\frac1{1+}\frac1{1+}\frac1{4+}\frac1{1+}\frac1{1+}\frac1{6+}\cdots\,, $$ I think you are mistaken in thinking that your number might even be algebraic, much less quadratic.

$\endgroup$
3
$\begingroup$

The continued fraction $0+\cfrac{1}{2+\cfrac{3}{4+\cfrac{5}{6+\cfrac{7}{8+\cfrac{9}{10+\ddots}}}}}$ can be evaluated using Euler's Differential method. Applying this method with $a=1, \alpha=2, b=2, \beta=2, c=1, \gamma=0$ results in a ODE. See Sergey Khruschev's Orthogonal Polynomials and Continued Fractions: From Euler's Point of View for technical details.
Solving the ODE results in $0+\cfrac{1}{2+\cfrac{3}{4+\cfrac{5}{6+\cfrac{7}{8+\cfrac{9}{10+\ddots}}}}}=\cfrac{\int_{0}^{1} e^{x/2}x^{1/2}dx}{\int_{0}^{1} e^{x/2}{x^{-1/2}}dx}=0.379732...$

Helas, this result can not be expressed easily in common functions. Of course, it can be expressed as the quotient of two series.

The interesting 'mirrored' Continued Fraction is $1+\cfrac{2}{3+\cfrac{4}{5+\cfrac{6}{7+\cfrac{8}{9+\cfrac{10}{11+\ddots}}}}}$

And this CF can also be evaluated using Euler's Differential method.

The interesting result is that $ 1+\cfrac{2}{3+\cfrac{4}{5+\cfrac{6}{7+\cfrac{8}{9+\cfrac{10}{11+\ddots}}}}}=\cfrac{1}{\sqrt{e} -1}=1.54149...$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .