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how to solve this equation:

$y^{\alpha} + y^{1 + \alpha} = x $ where $\alpha \in (-1, 0)$

Is there trick to solve it?

EDIT. I want to find $y(x)$.

ashim
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3 Answers3

4

Lagrange inversion formula, more precisely Lagrange-Bürmann formula is made for this.

The function $y$ solves $y=z\phi(y)$, with $z=x^{1/\alpha}$ and $\phi:t\mapsto(1+t)^{-1/\alpha}$. Since $\phi(0)\ne0$, one knows that a solution is $$ y(x)=\sum_{n\geqslant1}a_nz^n=\sum_{n\geqslant1}a_nx^{n/\alpha},\qquad a_n=\frac1n[t^{n-1}]\phi(t)^n. $$ In the present case, for every $\alpha$ in $(-1,0)$, $$ a_n=\frac1n{-n/\alpha\choose n-1}=-\frac{\alpha}n{-n/\alpha+1\choose n}, $$ and one can check that the resulting series converges for every $x$ such that $|x|\gt R_\alpha$ with $$ R_\alpha=\frac{(-\alpha)^\alpha}{(1+\alpha)^{1+\alpha}}. $$ Sanity check: If $\alpha=-\frac12$, one gets $R_\alpha=2$, as was to be expected since an explicit formula for $y(x)$ in this case is $$ y(x)=-1+\frac12x^2-\frac12x^2\sqrt{1-\frac4{x^2}}=\frac{1-\sqrt{1-\frac4{x^2}}}{1+\sqrt{1-\frac4{x^2}}}. $$ Edit: For every $\beta\gt0$ and $\gamma$, when $n\to\infty$, $$ \frac{\Gamma(\beta(n+1)+\gamma)}{\Gamma(\beta n+\gamma)}\sim \beta^\beta n^\beta. $$ Applying this to each Gamma function in the expression of $$ a_n=\dfrac{\Gamma(-n/\alpha+1)}{\Gamma(n+1)\Gamma(-n/\alpha-n+2)}, $$ one sees that the powers of $n$ cancel out and one is left with $$ \lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac{(-1/\alpha)^{-1/\alpha}}{1^1\,(-1/\alpha-1)^{-1/\alpha-1}}=(-1/\alpha)\,(1+\alpha)^{(1+\alpha)/\alpha}=\varrho(\alpha). $$ Hence the series $y(x)$ converges for every $x$ such that $\varrho(\alpha)|x|^{1/\alpha}\lt1$ and diverges for every $x$ such that $\varrho(\alpha)|x|^{1/\alpha}\gt1$. Since $\alpha\lt0$, the first condition reads $|x|\gt\varrho(\alpha)^{-\alpha}$ and the second condition reads $|x|\lt\varrho(\alpha)^{-\alpha}$. Since $\varrho(\alpha)^{-\alpha}=R_\alpha$, the proof is complete.

Did
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  • Your username got shorter again. Is the next step d or id? – Asaf Karagila May 31 '12 at 09:37
  • @Asaf: (i) Do you prefer one over the other? (ii) Do you know if there is a limit to the admissible length? (iii) Why not di? Oh... I see. – Did May 31 '12 at 09:40
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    "I have no idea" is the answer to all three; I suggest going for W if you choose to go one letter. – Asaf Karagila May 31 '12 at 10:08
  • @did could you explain please what is ${-n/\alpha\choose n-1}$ – ashim Jun 04 '12 at 04:32
  • See here. – Did Jun 04 '12 at 07:58
  • capoluca: Got something from the link indicated in my last comment? – Did Jun 07 '12 at 15:25
  • @did yes I did, sorry for late responce I just saw your comment. By the way how did you show convergence of the series when $|x|\gt R_\alpha$. thanks – ashim Jun 10 '12 at 16:29
  • The usual method: estimating the ratio $a_{n+1}/a_n$ when $n\to\infty$. – Did Jun 10 '12 at 16:40
  • @did could you explain a little bit more on your sanity check. in this case $a_n=\frac1n{2n\choose n-1}$, but how did you collaps infinit summ? – ashim Jun 10 '12 at 18:20
  • When $\alpha=-1/2$, the equation becomes $y-x\sqrt{y}+1=0$ hence $\sqrt{y}$ is an explicit function of $x$. Expanding the square yields the formula in my post, which is valid only when $4/x^2\leqslant1$. Thus the value $R=2$. – Did Jun 10 '12 at 19:08
  • @did could you clarify about convergence radius please. I if calculate ration $a_{n-1}/a_N$ then I cannot get rid of n :$\frac{(n-1)(-\frac n{\alpha} +1)\dots(-\frac n{\alpha} - (n-1))}{n^2 (-\frac{n-1}{\alpha} + 1)\dots(-\frac{n-1}{\alpha} - (n-2)))}$ – ashim Jun 11 '12 at 15:16
  • oh, sorry, I should take limit – ashim Jun 11 '12 at 15:23
  • See Edit. $ $ $ $ – Did Jun 11 '12 at 15:40
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Given that your class of equations includes polynomial equations of any degree, I suspect that a numerical approach would be needed, e.g. Newton-Raphson, and the value of $x$ would need to be specified. With a table of values of $x$, $y(x)$ could be constructed by interpolation. Notice that if we consider the case $\alpha=-1/n$ for a natural number $n>2$, the equation is equivalent to $z^n-xz+1=0$ where $z=y^{1/n}$.

user12477
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0

How about the parametric solution of parameter s?: $ x = \exp(s\alpha) [1+\exp(s)] $ and $y = \exp(s)$. You can plot these easily (I used Gnuplot). Note that (x,y) = (2,1) is always on the graph and note what happens when $s \to \pm\infty $. At least you get an idea about how y(x) looks like. Or you can use the Lagrange inversion formula around the fixed point to get an idea about the explicit y(x). Or you can invert the axes and look at it as x(y) to get the idea of how its plot looks like directly. Cheers!

Vasea Igor
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