Formula for the error of the nth convergent of the continued fraction representation of $E_1(x)$
By the continued fraction of the exponential integral,
$$E_1(x)=cfrac{e^{-x}}{x+cfrac{1}{1+cfrac{1}{x+cfrac{2}{1+cfrac{2}{cfrac{cdots}{x+cfrac{n}{1+cfrac{n}{x}}}}}}}}+R_n$$
where $R_n$ is a remainder.
I came up with the following formula.
$$R_n=frac{n+1}{-x L_{n}^{(1)}(-x)}sum_{k=1}^{n+1}{{n}choose{k-1}}(-1)^kE_{k+1}(x)$$
Or equivalently
$$R_n=frac{n+1}{-x L_{n}^{(1)}(-x)}int_{0}^1{{e^{-x/t}}(1-t)^ndt}$$
where $L_{k}^{(alpha)}(x)$ is the generalized Laguerre polynomial.
I figured this out a long time ago and now I can’t remember how I did it. Any help is appreciated.
reference-request continued-fractions
asked Apr 20 '17 at 4:12
tyobrien
1,176513
add a comment |
By the continued fraction of the exponential integral,
$$E_1(x)=cfrac{e^{-x}}{x+cfrac{1}{1+cfrac{1}{x+cfrac{2}{1+cfrac{2}{cfrac{cdots}{x+cfrac{n}{1+cfrac{n}{x}}}}}}}}+R_n$$
where $R_n$ is a remainder.
I came up with the following formula.
$$R_n=frac{n+1}{-x L_{n}^{(1)}(-x)}sum_{k=1}^{n+1}{{n}choose{k-1}}(-1)^kE_{k+1}(x)$$
Or equivalently
$$R_n=frac{n+1}{-x L_{n}^{(1)}(-x)}int_{0}^1{{e^{-x/t}}(1-t)^ndt}$$
where $L_{k}^{(alpha)}(x)$ is the generalized Laguerre polynomial.
I figured this out a long time ago and now I can’t remember how I did it. Any help is appreciated.
reference-request continued-fractions
asked Apr 20 '17 at 4:12
tyobrien
1,176513
add a comment |
By the continued fraction of the exponential integral,
$$E_1(x)=cfrac{e^{-x}}{x+cfrac{1}{1+cfrac{1}{x+cfrac{2}{1+cfrac{2}{cfrac{cdots}{x+cfrac{n}{1+cfrac{n}{x}}}}}}}}+R_n$$
where $R_n$ is a remainder.
I came up with the following formula.
$$R_n=frac{n+1}{-x L_{n}^{(1)}(-x)}sum_{k=1}^{n+1}{{n}choose{k-1}}(-1)^kE_{k+1}(x)$$
Or equivalently
$$R_n=frac{n+1}{-x L_{n}^{(1)}(-x)}int_{0}^1{{e^{-x/t}}(1-t)^ndt}$$
where $L_{k}^{(alpha)}(x)$ is the generalized Laguerre polynomial.
I figured this out a long time ago and now I can’t remember how I did it. Any help is appreciated.
reference-request continued-fractions
asked Apr 20 '17 at 4:12
tyobrien
1,176513
By the continued fraction of the exponential integral,
$$E_1(x)=cfrac{e^{-x}}{x+cfrac{1}{1+cfrac{1}{x+cfrac{2}{1+cfrac{2}{cfrac{cdots}{x+cfrac{n}{1+cfrac{n}{x}}}}}}}}+R_n$$
where $R_n$ is a remainder.
I came up with the following formula.
$$R_n=frac{n+1}{-x L_{n}^{(1)}(-x)}sum_{k=1}^{n+1}{{n}choose{k-1}}(-1)^kE_{k+1}(x)$$
Or equivalently
$$R_n=frac{n+1}{-x L_{n}^{(1)}(-x)}int_{0}^1{{e^{-x/t}}(1-t)^ndt}$$
where $L_{k}^{(alpha)}(x)$ is the generalized Laguerre polynomial.
I figured this out a long time ago and now I can’t remember how I did it. Any help is appreciated.
reference-request continued-fractions
reference-request continued-fractions
asked Apr 20 '17 at 4:12
tyobrien
1,176513
asked Apr 20 '17 at 4:12
tyobrien
1,176513
edited Dec 27 '18 at 6:30
asked Apr 20 '17 at 4:12
tyobrien
1,176513
asked Apr 20 '17 at 4:12
tyobrien
1,176513
asked Apr 20 '17 at 4:12
tyobrien
1,176513
1,176513
add a comment |
add a comment |
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