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.

  • $\begingroup$ Do you really want "powers of $\,z\,$" or in fact it is "as a function of $\,z\,$"? $\endgroup$ – DonAntonio Aug 1 '12 at 1:44
  • $\begingroup$ 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. $\endgroup$ – M. M. Aug 1 '12 at 2:16

As long as we remain within the convergence interval, and if you really wanted to express $\,y\,$ as a function of $\,z\,$ , we get the following differential equation:

$$y=x\sum_{k=0}^\infty a_kx^{2k}\Longrightarrow y'=\sum_{k=0}^\infty a_kx^{2k}+2\sum_{k=0}^\infty ka_kx^{2k}=\frac{y}{x}+\frac{2z}{x}=\frac{y+2z}{x}$$

  • $\begingroup$ The problem is actually to eliminate $x$ so this does not answer the question. Sorry, I just edited the question to make this clearer. $\endgroup$ – M. M. Aug 1 '12 at 2:09
  • $\begingroup$ @M.Mayrand , if you think something's wrong in my answer please do describe it here in a comment. Don't try to edit my answer anymore. $\endgroup$ – DonAntonio Aug 1 '12 at 2:42
  • $\begingroup$ The second sum in your expression for $y'$ is equal to $\frac{2}{x}z$. Hence the result is $\frac{y+2z}{x}$. $\endgroup$ – M. M. Aug 1 '12 at 2:52
  • $\begingroup$ The second sum in that expression for $\,y'\,$ is $$x\sum_{k=0}^\infty 2ka_kx^{2k-1}$$ with the usually accepted agreement that $\,w^r=0\,$ whenever $\,r<0\,$, but of course it doesn't mind: with the expression appearing in my answer (and not with the above one) there's no problem anymore, so the middle step can be done as you wish, but the result remains, imo, as when $\,k=0\,$ we get zero, so anyway things begin to count from $\,k=1\,$ . Open up the sumatories in both cases...Nevertheless there's indeed a typo: it must be $\,2/x\,$ and not $\,2/z\,$. It's been corrected, thank you. $\endgroup$ – DonAntonio Aug 1 '12 at 3:28
  • $\begingroup$ You are indeed right that the sum can start at k=0 (sorry I removed the comment before you answered but you missed it). However, I still see a typo: Where does the sum go? It must be 2z/x not 2/x. $\endgroup$ – M. M. Aug 1 '12 at 3:37

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