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edited Apr 11 2011 at 9:07 |
If we write the equation as $ax= (1+1/x)^{k-1}$ it is clear that there is exactly one positive solution (here and below I assume that $k$ is a real number larger than 1).
I. With the substitution $u=1/x$ the equation writes $a=u(1+u)^{k-1}$, and the RHS has a formal series inverse at $0$. The Lagrange inversion formula provides the (Laurent) expansion at $0$ of any integer power of the latter, in particular, of its reciprocal. One finds
$$x=a^{-1}+(k-1)+ \sum_{n=1}^\infty {(-1)^n \over n} {nk \choose n+1} a^n\ ,$$
according with the first terms found by Robert Israel.
This series has a radius of convergence $k^{-k}(k-1)^{k-1}$, so it gives the positive solution of the initial equation provided $a < k ^ {-k}(k-1) ^ {k-1} $. Moreover the terms of the series are eventually decreasing in absolute value, which implies that the $n$-th partial sums are eventually a bound on $x$ alternately from above and from below. If needed, one may also estimate the remainder from the partial sums using bounds on the binomial coefficients.
II. [edit] For large positive values of $a$, we can write the equation in the form
$$a^{-\frac{1}{k}}=x(1+x)^{-1+\frac{1}{k} } \ .$$
Again, the Lagrange inversion formula provides a power series expansion of the inverse of the RHS; computing the radius of convergence this time we find $k (k-1)^{\frac{1}{k}-1 }$. Hence, evaluating that inverse at $a^{-\frac{1}{k}}$ we get
$$x=\sum_{n=1}^\infty {1\over n}{\frac{n(k-1)}{k} \choose n-1 }a^{-{n\over k}}\ ,$$
converging for $a>k^{-k}(k-1)^{k-1}$. Again, estimates on the convergence are available with a little more work on the binomial coefficients. For $a=1/2$ and $k=5$, the first $20$ terms of the series sum to $x=4.4786$; compare with the numerical solution in Aleksey Pichugin's answer below.)
III. It remains the case $ a := k^{-k} (k-1)^{k-1} $, though it should be covered as a limit case by the above series.
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edited Apr 9 2011 at 8:00 |
If we write the equation as $ax= (1+1/x)^{k-1}$ it is clear that there is exactly one positive solution (here and below I assume that $k$ is a real number larger than 1).
I. With the substitution $u=1/x$ the equation writes $a=u(1+u)^{k-1}$, and the RHS has a formal series inverse at $0$. The Lagrange inversion formula provides the (Laurent) expansion at $0$ of any integer power of the latter, in particular, of its reciprocal. One finds
$$x=a^{-1}+(k-1)+ \sum_{n=1}^\infty {(-1)^n \over n} {nk \choose n+1} a^n\ ,$$
according with the first terms found by Robert Israel.
This series has a radius of convergence $k^{-k}(k-1)^{k-1}$, so it gives the positive solution of the initial equation provided $a < k ^ {-k}(k-1) ^ {k-1} $. Moreover the terms of the series are eventually decreasing in absolute value, which implies that the $n$-th partial sums are eventually a bound on $x$ alternately from above and from below. If needed, one may also estimate the remainder from the partial sums using bounds on the binomial coefficients.
II. [edit] For large positive values of $a$, one we can write the equation in the form
$$a^{-\frac{1}{k}}=x(1+x)^{-1+\frac{1}{k} } \ .$$
Again, the Lagrange inversion formula provides a power series expansion of the RHS; computing the radius of convergence this time one finds we find $k (k-1)^{\frac{1}{k}-1 }$. Hence, evaluating the that inverse at $a^{-\frac{1}{k}}$ we get
$$x=\sum_{n=1}^\infty {1\over n}{\frac{n(k-1)}{k} \choose n-1 }a^{-{n\over k}}\ ,$$
converging for $a>k^{-k}(k-1)^{k-1}$. Again, estimates on the convergence are available with a little more work on the binomial coefficients; anyway now the convergence is quite fast, due to the decay of the binomial coefficients. For $a=1/2$ and $k=5$, the first $20$ terms of the series sum to $x=4.4786$; compare with the numerical solution in Aleksey Pichugin's answer below.)
III. It remains therefore the case $ a _ 0 := k^{-k} (k-1)^{k-1} $; although , though it should be covered as a limit case by the above series.
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edited Apr 9 2011 at 3:10 |
If we write the equation as $ax= (1+1/x)^{k-1}$ it is clear that there is exactly one positive solution (here and below I assume that $k$ is a real number larger than 1).
I. With the substitution $u=1/x$ the equation writes $a=u(1+u)^{k-1}$, and the RHS has a formal series inverse at $0$. The Lagrange inversion formula provides the (Laurent) expansion at $0$ of any integer power of the latter, in particular, of its reciprocal. One finds
$$x=a^{-1}+(k-1)+ \sum_{n=1}^\infty {(-1)^n \over n} {nk \choose n+1} a^n\ ,$$
according with the first terms found by Robert Israel.
This series has a radius of convergence $k^{-k}(k-1)^{k-1}$, so it gives the positive solution of the initial equation provided $a
II. [edit] For large positive values of $a$, one can write the equation in the form
$$a^{-\frac{1}{k}}=x(1+x)^{-1+\frac{1}{k} } \ .$$
Again, the Lagrange inversion formula provides a power series expansion of the RHS; computing the radius of convergence this time one finds $k (k-1)^{\frac{1}{k}-1 }$. Hence, evaluating the inverse at $a^{-\frac{1}{k}}$
$$x=\sum_{k=1}^\infty $x=\sum_{n=1}^\infty {1\over n}{\frac{n(k-1)}{k} \choose n-1 }a^{-{n\over k}}\ ,$$
converging for $a>k^{-k}(k-1)^{k-1}$. Again, estimates on the convergence are available with a little more work on the binomial coefficients; anyway now the convergence is quite fast, due to the decay of the binomial coefficients. For $a=1/2$ and $k=5$, the first $20$ terms of the series sum to $x=4.4786$; compare with the numerical solution in Aleksey Pichugin's answer below.)
III. It remains therefore the case $ a _ 0 := k^{-k} (k-1)^{k-1} $; although it should be covered as a limit case by the above series, it could be nice to see a power series solution for values of $a$ in a neighbourhood of $a_0$..
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edited Apr 8 2011 at 20:31 |
If we write the equation as $ax= (1+1/x)^{k-1}$ it is clear that there is exactly one positive solution (here and below I assume that $k$ is a real number larger than 1).
I. With the substitution $u=1/x$ the equation writes $a=u(1+u)^{k-1}$, and the RHS has a formal series inverse at $0$. The Lagrange inversion formula provides the (Laurent) expansion at $0$ of any integer power of the latter, in particular, of its reciprocal. One finds
$$x=a^{-1}+(k-1)+ \sum_{n=1}^\infty {(-1)^n \over n} {nk \choose n+1} a^n\ ,$$
according with the first terms found by Robert Israel.
This series has a radius of convergence $k^{-k}(k-1)^{k-1}$, so it gives the positive solution of the initial equation provided $a
II. [edit] For large positive values of $a$, one can write the equation in the form
$$a^{-\frac{1}{k}}=x(1+x)^{-1+\frac{1}{k} } \ .$$
Again, the Lagrange inversion formula provides a power series expansion of the RHS; computing the radius of convergence this time one finds $k (k-1)^{\frac{1}{k}-1 }$. Hence, evaluating the inverse at $a^{-\frac{1}{k}}$
$$x=\sum_{k=1}^\infty {1\over n}{\frac{n(k-1)}{k} \choose n-1 }a^{-{n\over k}}\ ,$$
converging for $a>k^{-k}(k-1)^{k-1}$. Again, estimates on the convergence are available with a little more work on the binomial coefficients; anywa ynow anyway now the convergence is quite fast, due to the decay of the binomial coefficients)coefficients. For $a=1/2$ and $k=5$, the first $20$ terms of the series sum to $x=4.4786$; compare with the numerical solution in Aleksey Pichugin's answer below.)
III. It remains therefore the case $ a _ 0 := k^{-k} (k-1)^{k-1} $; although it should be covered as a limit case by the above series, it could be nice to see a power series solution for value values of a $a$ in a neighbourhood of $a_0$.
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edited Apr 8 2011 at 19:38 |
If we write the equation as $ax= (1+1/x)^{k-1}$ it is clear that there is exactly one positive solution (here and below I assume that $k$ is a real number larger than 1).
I. With the substitution $u=1/x$ the equation writes $a=u(1+u)^{k-1}$, and the RHS has a formal series inverse at $0$. The Lagrange inversion formula provides the (Laurent) expansion at $0$ of any integer power of the latter, in particular, of its reciprocal. One finds
$$x=a^{-1}+(k-1)+ \sum_{n=1}^\infty {(-1)^n \over n} {nk \choose n+1} a^n\ ,$$
according with the first terms found by Robert Israel.
This series has a radius of convergence $k^{-k}(k-1)^{k-1}$, so it gives the positive solution of the initial equation provided $a
II. [edit] For large positive values of $a$, one can write the equation in the form
$$a^{-\frac{1}{k}}=x(1+x)^{-1+\frac{1}{k} } \ .$$
Again, the Lagrange inversion formula provides a power series expansion of the RHS; computing the radius of convergence this time one finds $k (k-1)^{\frac{1}{k}-1 }$. Hence, computing evaluating the inverse at $a^{-\frac{1}{k}}$
$$x=\sum_{k=1}^\infty {1\over n}{\frac{n(k-1)}{k} \choose n-1 }a^{-{n\over k}}\ ,$$
converging for $a>k^{-k}(k-1)^{k-1}$. Again, estimates on the convergence are available with a little more work on the binomial coefficients; anyway now anywa ynow the convergence is quite fast due to the decay of the binomial coefficients).
III. It remains therefore the case $ a _ 0 := k^{-k} (k-1)^{k-1} $; although it should be covered as a limit case by the above series, it could be nice to see a power series solution for valuse value of a in a neighbourhood of $a_0$.
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edited Apr 8 2011 at 18:46 |
If we write the equation as $ax= (1+1/x)^{k-1}$ it is clear that there is exactly one positive solution (here and below I assume that $k$ is a real number larger than 1).
I. With the substitution $u=1/x$ the equation writes $a=u(1+u)^{k-1}$, and the RHS has a formal series inverse at $0$. The Lagrange inversion formula provides the (Laurent) expansion at $0$ of any integer power of the latter, in particular, of its reciprocal. One finds
$$x=a^{-1}+(k-1)+ \sum_{n=1}^\infty {(-1)^n \over n} {nk \choose n+1} a^n\ ,$$
according with the first terms found by Robert Israel.
This series has a radius of convergence $R_k:=k^{-k}(k-1)^{k-1}$, k^{-k}(k-1)^{k-1}$, so it gives the positive solution of the initial equation provided $a
< R_k$ II. Moreover the terms [edit] For large positive values of $a$, one can write the series are eventually decreasing equation in absolute value, which implies that the $n$-th partial sums are eventually a bound on form
$x$ alternately from above and from below$a^{-\frac{1}{k}}=x(1+x)^{-1+\frac{1}{k} } \ .If needed$$
Again, the Lagrange inversion formula provides a power series expansion of the RHS; computing the radius of convergence this time one may also estimate finds $k (k-1)^{\frac{1}{k}-1 }$. Hence, computing the remainder from inverse at $a^{-\frac{1}{k}}$
$$x=\sum_{k=1}^\infty {1\over n}{\frac{n(k-1)}{k} \choose n-1 }a^{-{n\over k}}\ ,$$
converging for $a>k^{-k}(k-1)^{k-1}$. Again, estimates on the partial sums using bounds convergence are available with a little more work on the binomial coefficients; anyway now the convergence is quite fast due to the decay of the binomial coefficients).
III. It remains therefore the case $ a _ 0 := k^{-k} (k-1)^{k-1} $; although it should be covered as a limit case by the above series, it could be nice to see a power series solution for valuse of a in a neighbourhood of $a_0$.
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edited Apr 7 2011 at 22:13 |
If we write the equation as $ax= (1+1/x)^{k-1}$ it is clear that there is exactly one positive solution (here and below I assume that $k$ is a real number larger than 1).
With the substitution $u=1/x$ the equation writes $a=u(1+u)^{k-1}$, and the RHS has a formal series inverse at $0$. The Lagrange inversion formula provides the (Laurent) expansion at $0$ of any integer power of the latter, in particular, of its reciprocal. One finds
$$x=a^{-1}+(k-1)+ \sum_{n=1}^\infty {(-1)^n \over n} {nk \choose n+1} a^n\ ,$$
according with the first terms found by Robert Israel.
This series has a radius of convergence $R_k:=k^{-k}(k-1)^{k-1}$, so it gives the positive solution of the initial equation provided $a < R_k$. Moreover the terms of the series are eventually decreasing in absolute value, which implies that the $n$-th partial sums are eventually a bound on $x$ alternately from above and from below$x$. . If needed, one may also estimate the remainder from the partial sums using bounds on the binomial coefficients.
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answered Apr 7 2011 at 21:31 |
If we write the equation as $ax= (1+1/x)^{k-1}$ it is clear that there is exactly one positive solution (here and below I assume that $k$ is a real number larger than 1).
With the substitution $u=1/x$ the equation writes $a=u(1+u)^{k-1}$, and the RHS has a formal series inverse at $0$. The Lagrange inversion formula provides the (Laurent) expansion at $0$ of any integer power of the latter, in particular, of its reciprocal. One finds
$$x=a^{-1}+(k-1)+ \sum_{n=1}^\infty {(-1)^n \over n} {nk \choose n+1} a^n\ ,$$
according with the first terms found by Robert Israel.
This series has a radius of convergence $R_k:=k^{-k}(k-1)^{k-1}$, so it gives the positive solution of the initial equation provided $a < R_k$. Moreover the terms of the series are eventually decreasing in absolute value, which implies that the $n$-th partial sums are eventually a bound on $x$ alternately from above and from below $x$. If needed, one may also estimate the remainder from the partial sums using bounds on the binomial coefficients.
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