Continued fractions can be represented by the numerator of the nth convergent divided by the denominator of the nth convergent. (See "some useful theorems" on continued fractions). I tried this with the exponential integral. $$\mathrm{Ei}(x)=\cfrac{e^x}{x+\cfrac{1}{-1+\cfrac{1}{x+\cfrac{2}{-1+\cfrac{2}{x+\cfrac{3}{\cdots}}}}}}=\cfrac{e^x}{x+\cfrac{1}{-1+\cfrac{1}{x+\cfrac{1}{-1/2+\cfrac{1}{x+\cfrac{1}{-1/3+\cfrac{1}{\cdots}}}}}}}$$ Through this question and some work on my own, I got that the Nth convergents of the continued fraction that starts under the exponential are $$\frac{\frac{1}{N!}\sum_{k=0}^N{(-1)}^k\left(\prod_{i=1}^{N-k}\frac{(i+k)(i+k+1)}{i}\right)x^{k+1}}{1-\frac{1}{N!}e^{-x}\sum_{k=1}^N{(-1)}^k\left(\prod_{i=1}^{N-k}\frac{(i+k)(i+k+1)}{i}\right)x^{k+1}\sum_{n=1}^k(-1)^n{k\choose n}{\frac{d^n}{dx^n}\mathrm{Ei}(x)}}$$ So flipping the fraction and multiplying by $e^x$, we have $$\mathrm{Ei}(x)=\lim\limits_{N\to\infty}\frac{e^x-\frac{1}{N!}\sum_{k=1}^N{(-1)}^k\left(\prod_{i=1}^{N-k}\frac{(i+k)(i+k+1)}{i}\right)x^{k+1}\sum_{n=1}^k(-1)^n{k\choose n}{\frac{d^n}{dx^n}\mathrm{Ei}(x)}}{\frac{1}{N!}\sum_{k=0}^N{(-1)}^k\left(\prod_{i=1}^{N-k}\frac{(i+k)(i+k+1)}{i}\right)x^{k+1}}$$ Using continued fractions to find a new formula could be done with other functions as well, but I found it especially interesting that here we have the function in terms of its derivatives. Is it a coincidence that this happens? If not, does it mean that we can find formulas for other functions in terms of their derivatives using this continued fraction method?

In addition, If this exponential integral expression can be simplified, please show.

UPDATE: This formula also appears to work for any exponential function, $c^x$; $x\lt 0$, not just the exponential integral. So to narrow my question, for what functions does this work and why? And how might it be useful?

UPDATE: A slightly better-looking version is $$\mathrm{Ei}(x)=\lim\limits_{N\to\infty}\frac{e^x-\frac{1}{N!}\sum_{k=1}^N{(-1)}^k{L(N+1,k+1)}x^{k+1}\sum_{n=1}^k(-1)^n{k\choose n}{\frac{d^n}{dx^n}\mathrm{Ei}(x)}}{\frac{1}{N!}\sum_{k=0}^N{(-1)}^k{L(N+1,k+1)}x^{k+1}}$$ Where $L(n,k)={{n-1}\choose{k-1}}\frac{n!}{k!}$ are the Lah numbers.

Now the denominator can be written in terms of Laguerre Polynomials. $$\mathrm{Ei}(x)=\lim\limits_{N\to\infty}\frac{e^x-\frac{1}{N!}\sum_{k=1}^N{(-1)}^k{L(N+1,k+1)}x^{k+1}\sum_{n=1}^k(-1)^n{k\choose n}{\frac{d^n}{dx^n}\mathrm{Ei}(x)}}{x{L_{N}^{(1)}}(x)}$$ Now if only the top could be simplified in terms of Laguerre Polynomials or something to avoid confusion.

UPDATE: I actually was able to simplify it much more and get this result.

  • $\begingroup$ Are you aware of Thiele continued fractions, by any chance? $\endgroup$ – J. M. is a poor mathematician Apr 16 '17 at 0:43
  • $\begingroup$ No I am not @J.M.isn'tamathematician $\endgroup$ – tyobrien Apr 16 '17 at 0:50

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.