solving a problem with generating functions

Question

This is a problem from a course of MIT.

Find the coefficients of the power series $y = 1 + 3 x + 15 x^2 + 184 x^3 + 495 x^4 + \cdots $ satisfying

$$ (27 x - 4)y^3 + 3y + 1 = 0 . $$

This is an interesting problem, but I am really cluelss (My naive approaches all failed). Can anyone give me a hint? Or an answer?

Answer 1

Too long for a comment, not really an answer:

Let's work with formal power series.

Let $y=\sum a_nx^n$ with $a_0=1,a_1=3,a_2=15,a_3=184,a_4=495$. Then knowing that $y$ satisfies the given equation we know that $$(27x-4)\left(\sum a_nx^n\right)^3+3\sum a_nx^n+1=0=\sum0\cdot x^n$$ In order to compare coefficients we need to express $\left(\sum a_nx^n\right)^3=\sum b_nx^n$. There are various ways but a possible one is $$\sum_n\left(\sum_{i,j,k\\ i+j+k=n}a_ia_ja_k\right)x^n$$ from which you get the recursive equation \begin{align*}a_0=1,a_1=3,a_2=15,a_3=184,a_4=495\\ 27\sum_{i,j,k\\ i+j+k=n-1}a_ia_ja_k-4\sum_{i,j,k\\ i+j+k=n}a_ia_ja_k+3a_n=0\quad n\geq 5\end{align*} where in the last expression we can solve for $a_n$ in order to calculate this term using the previous (know) coefficient (using that $a_0=1$) $$27\sum_{i,j,k\\ i+j+k=n-1}a_ia_ja_k-4\cdot 3\cdot a_n-4\sum_{i,j,k\neq n\\ i+j+k=n}a_ia_ja_k+3a_n=0$$ from which we get $$a_n=\frac{1}{9}\left(27\sum_{i,j,k\\ i+j+k=n-1}a_ia_ja_k-4\sum_{i,j,k\neq n\\ i+j+k=n}a_ia_ja_k\right)$$

Answer 2

I think that you're supposed to use the equation and assuming that $y$ is given by a power series, prove that the first five terms are the ones that are given.

By solving the third degree equation it can be shown that the solution that approaches $1$ when $x\rightarrow 0$ is $$ y = \frac{\cos(\frac13\arcsin\sqrt{\frac{27x}{4}})}{\sqrt{1-\frac{27x}{4}}} $$ I doubt that the general coefficients of its Taylor series can be expressed in some nice way.

To derive these first five coefficients and be able to check them, you put $$y = \sum_{n=0}^4 a_0 x^n + O(x^5)$$ into your equation to obtain \begin{align} & (-4a_0^3 +3a_0 + 1) + (-12a_0^2a_1 + 27a_0^3 + 3a_1 ) x + (-12a_0^2a_2-12a_0a_1^2+81a_0^2a_1+3a_2)x^2 + \\ &\quad + (-12a_0^2a_3-24a_0a_1a_2-4a_1^3+81a_0^2a_2+81a_0a_1^2+3a_3)x^3 + \\ &\quad + (-12a_0^2a_4-24a_0a_1a_3 - 12a_0a_2^2 - 12a_1^2a_2 +81a_0^2a_3+162a_0a_1a_2+3a_4+27a_1^3)x^4 + \\ &\quad + O(x^5) = 0 \end{align} That is $$\left\{\begin{array}{l} -(a_0-1)(2a_0+1)^2 = 0 \\ a_1(3-12a_0^2) + 27a_0^3 = 0 \\ a_2(3-12a_0^2) -12a_0a_1^2+81a_0^2a_1 = 0\\ a_3(3-12a_0^2)-24a_0a_1a_2-4a_1^3+81a_0^2a_2+81a_0a_1^2 = 0 \\ a_4(3-12a_0^2)-24a_0a_1a_3 - 12a_0a_2^2 - 12a_1^2a_2 +81a_0^2a_3+162a_0a_1a_2+27a_1^3 = 0\end{array}\right.$$ The first equation has two solutions, $a_0=1$ and $a_0=-\frac12$, but for $a_0=-\frac12$ the subsequent equations don't have solutions. If you start with $a_0=1$, then you can recursively obtain subsequent coefficients.

Answer 3

Much inspired by @Adam Latosiński's answer.

If we consider the depressed cubic equation $$(27 x - 4)y^3 + 3y + 1 = 0 $$ we have $$\Delta=-729 ( 27 x-4)x$$ which is negative if $x <0$ or $x > \frac 4{27}$.

On the other hand $$p=\frac{3}{27 x-4}\qquad \text{and} \qquad q=\frac{1}{27 x-4}=\frac p 3$$ If we have a single root with $\Delta<0$ and $p <0$ (since we are looking for expansions around $x=0$, the solution given by the hyperbolic method $$y=-\frac{2 \sqrt{-p} |p| }{\sqrt{3} p}\cosh \left(\frac{1}{3} \cosh ^{-1}\left(-\frac{\sqrt{3} \sqrt{-\frac{1}{p}} |p|}{2 p}\right)\right)$$ that is to say $$\color{blue}{y=\frac{2 \cosh \left(\frac{1}{3} \cosh ^{-1}\left(\frac{1}{2} \sqrt{4-27 x}\right)\right)}{\sqrt{4-27 x}}}$$ Useless but "amazing", let $t=\frac{1}{2} \sqrt{4-27 x}$ to make $$y=\frac{\cosh \left(\frac{1}{3} \cosh ^{-1}(t)\right)}{t}$$ and compose Taylor series around $t=1$ to get $$y=1-\frac{8 (t-1)}{9}+\frac{212}{243} (t-1)^2-\frac{5696 (t-1)^3}{6561}+\frac{51184 (t-1)^4}{59049}-\frac{4143616 (t-1)^5}{4782969}+\frac{111854336 (t-1)^6}{129140163}-\frac{1006606336 (t-1)^7}{1162261467}+\frac{27177461504 (t-1)^8}{31381059609}+O\left((t-1)^9\right)$$ and back to $x$ $$y=1+3 x+15 x^2+84 x^3+495 x^4+3003 x^5+18564 x^6+116280 x^7+735471 x^8+O\left(x^9\right)$$ which is simply the beginning of $$\color{blue}{y=\sum_{n=0}^\infty \binom{3 n}{n} x^n}$$

In fact, being lazy and typing in $OEIS$ the sequence $\{1, 3, 15,\color{red}{?}, 495\}$, one of the five found sequences is $A005809$ which corresponds to $\binom{3 n}{n}$. In the documentation, Tom Copeland reports the generating function given by @Adam Latosiński.

Answer 4

The given coefficients of the power series \begin{align*} y(x)=1+3x+15x^2+\color{blue}{84}x^3+495x^4+\cdots \end{align*} indicate they are $\binom{3n}{n}$ resulting in \begin{align*} y(x)=\sum_{n=0}^\infty \binom{3n}{n}x^n \tag{1} \end{align*}

We prove (1) by showing it fulfils the functional equation \begin{align*} (27 x-4)y^3+3y+1=0\tag{2} \end{align*}

We do so by emplyoing the Lagrange inversion following the paper Lagrange Inversion: when and how by R. Sprugnoli (et al.). We also use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.

We note the coefficients $\binom{3n}{n}$ from (1) are the coefficients of \begin{align*} \binom{3n}{n}=[x^n](1+x)^{3n}\tag{3} \end{align*}

Let us suppose that a formal power series $w=w(x)$ is implicitely defined by a relation $w=x\Phi(w)$, where $\Phi(x)$ is a formal power series such that $\Phi(0)\ne0$. The Lagrange Inversion Formula (LIF) states that:

$$[x^n]w(x)^k=\frac{k}{n}[x^{n-k}]\Phi(x)^n$$

There are several variations of the LIF stated in the paper. We use in the following formula $G6$ (with $F(x)=1)$:

Let $w=x\Phi(w)$ as before, then the following is valid:

\begin{align*} [x^n]\Phi(x)^n=\left[\left.\frac{1}{1-x\Phi'(w)}\right|w=x\Phi(w)\right]\tag{4} \end{align*}

The notation $[\left.f(w)\right|w=g(x)]$ is a linearization of $\left.f(w)\right|_{w=g(x)}$ and denotes the substitution of $g(x)$ to each occurrence of $w$ in $f(w)$ (that is, $f(g(x))$). In particular, $w=x\Phi(w)$ is to be solved in $w=w(x)$ and $w$ has to be substituted in the expression on the left of the $|$ sign.

We obtain from (4) with $\Phi(x)=(1+x)^3$ and using \begin{align*} x\Phi^{\prime}(w)=3x(1+w)^2=\frac{3x\Phi(w)}{1+w}=\frac{3w}{1+w} \end{align*} \begin{align*} \binom{3n}{n}&=[x^n](1+x)^{3n}\\ &=[x^n]\left[\left.\frac{1}{1-x\Phi^{\prime}(w)}\right|w=x\Phi(w)\right]\\ &=[x^n]\left[\left.\frac{1}{1-\frac{3w}{1+w}}\right|w=x\Phi(w)\right]\\ &=[x^n]\left[\left.\frac{1+w}{1-2w}\right|w=x\Phi(w)\right]\tag{5}\\ \end{align*}

It follows from (5) \begin{align*} y(x)=\sum_{n=0}^\infty\binom{3n}{n}x^n=\frac{1+w(x)}{1-2w(x)} \qquad\text{resp.}\qquad w(x)=\frac{y(x)-1}{2y(x)+1} \end{align*}

Since $w=x\Phi(w)=x(1+w)^3$, we finally obtain \begin{align*} \frac{y(x)-1}{2y(x)+1)}=x\left(1+\frac{y(x)-1}{2y(x)+1}\right)^3 \end{align*} from which \begin{align*} \color{blue}{(27x-4)y(x)^3+3y(x)+1=0} \end{align*} follows, showing the relationship (2) is fulfilled by (1).