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Suppose that the sequence $\{a_i\}_{i=0}^\infty$ is known and that, $$ y=\sum_{k=0}^{\infty}a_k x^{2k+1},\quad\text{and}\quad z=\sum_{k=1}^{\infty}k a_k x^{2k+1} $$ are two convergent power series for at least a finite radius of convergence.

What is the simplest way to eliminate $x$ and find a series for $y$ in powers of $z$?

The series has the form, $$ y=\sum_{k=0}^{\infty}b_k z^{\frac{2k+1}{3}} $$ so the question is to find the $b_k$s in terms of the $a_i$s. (The convergence of this series is not part of the question.)

I found a way to solve it but it is very complicated, so I am asking if someone knows a simpler way or is able to reduce the number of recurrence relations in my result. For example, is there a way to find the $b_i$s with a simple use of Bell's polynomials?

To solve the problem, I first took the cube root of $z$ to get a series for $z^{1/3}$ where the coefficients are then found by a recursive formula. Then I inversed $z$ to get a series for $x$ in powers of $z$ which adds another recursive formula. Then I substituted in the series for $y$ and expanded it which adds yet another recursive formula. Here is the result : $$ b_k=\sum_{i=0}^k a_i c_{k-i}^{(2i+1)} $$ Where, $$ c_0^{(k)}=\frac{1}{a_1^{k/3}},\quad c_n^{(k)}=a_1^{1/3}\sum_{i=1}^{n}\left(\frac{i(k+1)}{n}-1\right)u_i c_{n-i}^{(k)} $$ $$ u_0=\frac{1}{a_1^{1/3}},\quad u_n=-\frac{1}{a_1^{1/3}}\sum_{i=1}^n v_i c_{n-i}^{(2i+1)} $$ $$ v_0=a_1^{1/3},\quad v_n=\frac{1}{a_1}\sum_{i=1}^{n}\left(\frac{4i}{3n}-1\right)(i+1)a_{i+1} v_{n-i} $$ Hence, the first few terms are $$ y=\frac{1}{a_1^{1/3}}z^{1/3}-\frac{2a_2}{3a_1^2}z+\frac{16a_2^2-9a_1a_3}{9a_1^{11/3}}z^{5/3}-\frac{520 a_2^3-540 a_1 a_2 a_3+108 a_1^2 a_4}{81 a_1^{16/3}}z^{7/3}+\cdots $$

I verified with Mathematica using the Series and InverseSeries functions that this is the correct answer.

  • 0
    Do you *really* want "powers of $\,z\,$" or in fact it is "as a function of $\,z\,$"?2012-08-01
  • 0
    I want any formula that gives a value for $y$ by giving a value for $z$. I am quite sure that the only possibility is a power series.2012-08-01

1 Answers 1