Expanding f(z) with Known Moments: m_k

[zetafunction]

given the set of moments $$m_k$$

obtained from the known measure w(x) $$\int _{-\infty}^{\infty}dx w(x) x^{k} = m_k$$

now if we define the moment generating function $$f(z)= \sum_{k=0}^{\infty} m_{k}(-1)^{k}z^{k}$$ , which have the integral representation

$$f(z)= \int _{-\infty}^{\infty} \frac{w(x)}{1+xz }$$

my question is , how could i expand f(z) knowing the moment m_k in the form of a continued fraction

$$f(z) = a_0 (z)+ \frac{1 \mid}{\mid a_1 (z)} + \frac{1 \mid}{\mid a_2 (z)} + \frac{1 \mid}{\mid a_3 (z)}.$$

where the a_n are Polynomials of degree 1 .

for example for the case of exponential integral , how did euler manage to get the continued fraction ?? thanks.

 

[jvicens]

Thank you for your question. The expansion of the moment generating function f(z) in the form of a continued fraction can be obtained using the Cauchy integral formula. This formula states that for a function f(z) that is analytic in a region containing a closed contour C and a point z inside C, the coefficient of z^n in the Taylor series expansion of f(z) about z=0 is given by:

\frac{1}{2\pi i}\oint_C \frac{f(z)}{z^{n+1}} dz

In our case, we can use this formula to obtain the coefficients of z^n in the Taylor series expansion of f(z) by choosing the contour C to be a circle centered at z=0 with radius R large enough to contain all the singularities of f(z). Then, by expanding the fraction \frac{1}{1+xz} in a geometric series, we can express f(z) as a sum of terms of the form \frac{w(x)}{x^{n+1}}. Using the Cauchy integral formula, we can then obtain the coefficients of z^n in the Taylor series expansion of f(z) as integrals of the form \int_{-\infty}^{\infty} \frac{w(x)}{x^{n+1}} dx, which are just the moments m_n. Therefore, we have:

f(z) = \sum_{n=0}^{\infty} m_n (-1)^n z^n

which is the desired continued fraction expansion.

As for the specific case of the exponential integral, Euler was able to obtain the continued fraction expansion by using the Maclaurin series expansion of the exponential function e^z and the definition of the exponential integral as a logarithmic integral. By manipulating these series and integral expressions, he was able to obtain the continued fraction representation for the exponential integral.

I hope this helps to answer your question. If you have any further questions, please don't hesitate to ask.