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How to prove that this continued fraction $1+\cfrac{1/1}{1+\cfrac{1/3}{1+\cfrac{1/5}{1+\cfrac{1/7}{1+\ddots}}}}$ evaluates to $\displaystyle\frac{2}{\displaystyle 1+\frac{I_1(\frac14)}{I_0(\frac14)}}$?

The recurrence relation of these Besselfunctions is a different one than the recurrence associated with this Continued Fraction.

Paul vdVeen
  • 929

1 Answers1

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The continued fraction $\cfrac{1}{1+\cfrac{1/1}{1+\cfrac{1/3}{1+\cfrac{1/5}{1+\cfrac{1/7}{1+\ddots}}}}}$ can be evaluated as follows:

Consider $w_n=\cfrac{1}{1+\cfrac{w_{n+1}}{(2n+1)}}$ Assume $w_n=\cfrac{I_{n+1}}{I_n}$

This results in the recurrence relation ${(1+2n)I_n}={(1+2n)I_{n+1}}+{I_{n+2}}$

Euler's Differential method can be used to find an integral which satisfies this recurrence. Applying Euler's method with $a=1, \alpha=2, b=1, \beta=2, c=1, \gamma=0$ results in a ODE. See Sergey Khruschev's Orthogonal Polynomials and Continued Fractions: From Euler's Point of View for technical details of Euler's Differential Method. The choice R(x)=x satisfies and results in: $$w_0=\cfrac{I_1}{I_0}=\cfrac{1}{1+\cfrac{1/1}{1+\cfrac{1/3}{1+\cfrac{1/5}{1+\ddots}}}}=\cfrac{\int_{0}^{1} e^{x/2}\cfrac{x^{1/2}}{(x-1)^{1/2}}dx}{\int_{0}^{1} e^{x/2}\cfrac{1}{x^{1/2}(x-1)^{1/2}}dx}$$

Using the substitution $x=\cfrac{1+cos(\phi)}{2}$ it follows easily that this quotient of integrals equals $\frac{\displaystyle I_0(\frac14)+I_1(\frac14)}{\displaystyle2I_0(\frac14)}$

Taken reciprocals proves that $1+\cfrac{1/1}{1+\cfrac{1/3}{1+\cfrac{1/5}{1+\cfrac{1/7}{1+\ddots}}}} = \frac{\displaystyle 2}{\displaystyle 1+\frac{\displaystyle I_1(\frac14)}{I_0(\frac14)}}$

Paul vdVeen
  • 929
  • What is $,I_n,$ in the recurrence? What is the initial conditions of the recurrence? I can guess, but you should be explicit. What is Euler's Differential method? What is a DV? How does the substitution in integrals result in Bessel functions? Please be more explicit and supply the missing details. – Somos Oct 11 '21 at 21:18
  • Euler's Differential Method is a way to convert a Continued Fraction into a recurrence relation and to a Differential Equation (I used DV because in Dutch expression is "Differential Vergelijking") Euler's approach results in a quotient of two integrals as the outcome of the Continued Fraction. Solving this DE results in a quotient of two integrals. See for a very clear explanation the articles of Sergey Khrushchev "Orthogonal-polynomials-and-continued-fractions" Using the substitution, these integrals can easily be converted to a quotient of Modified Bessel functions. – Paul vdVeen Oct 12 '21 at 22:32
  • Sorry, but you still have not answered all of my questions. What is $,I_n,$ in the recurrence? Why can't you add the definition of $,I_n,$ to your answer? What is $,R(x),$ supposed to be? Do you have a link to somewhere where these are defined? How does the substitution lead to Bessel functions? Please give the answers to these questions in your answer. – Somos Oct 12 '21 at 23:02
  • Continued Fractions can be converted in a standard way to a recurrence relation. Euler's Differental Method is a method to find integrals which satisfies the recurrence. In this way the Continued Fraction is evaluated as a quotient of two integrals. When a simple substitution is applied, these integrals appear to be modified Bessel functions. As often in mathematics, the margins do not allow to write down the entire calculation. In evaluating these integrals WolframMathematics can be of use. (and will show the same resulting quotient of Besselfunctions) – Paul vdVeen Oct 13 '21 at 08:47