I am dealing with this...


Given $y$ is a function of $x$, $x^n+y^n=1$, where $n$ is a positive integer.

Find $\displaystyle\frac{d^ny}{dx^n}$ in terms of $x$, $y$ and $n$.

Example 1

For example, when $n=1$, $x+y=1$ then


Example 2

When $n=3$, $x^3+y^3=1$ then


Example 3

When $n=6$, $x^6+y^6=1$ then

$\displaystyle\frac{d^6y}{dx^6}=-\left(\frac{623645x^{30}}{y^{35}}+\frac{1612875x^{24}}{y^{29}}+\frac{1425875x^{18}}{y^{23}}+\frac{482625x^{12}}{y^{17}}+\frac{46100x^6}{y^{11}}+\frac{120}{y^5} \right)$.

Table of values

Below is a triangular array of values for the coefficient:

\begin{array}{|c|c|c|c|} \hline n \backslash k& 1 & 2 & 3 & 4 & 5 & 6 &.\\ \hline 1 & 1\\ \hline 2 & 1 & 1&\\ \hline 3 & 10& 12&2&\\ \hline 4 & 231&378&153&6 &\\ \hline 5 & 9576&20160&12960&2400 & 24&\\ \hline 6 & 623645&1612875 & 1425875&482625&46100&120&\\ \hline. \end{array}

Denote $a_{n,k}$ (where $k$ is a positive integer $\le n$) as the number in the $n$-th row and the $k$-th column. (e.g. $a_{3,2}=12$)

\begin{align} \displaystyle \frac{d^ny}{dx^n}=-\sum_{k=1}^n a_{n,k}\left(\frac{x^{n^2-kn}}{y^{n^2-kn+n-1}}\right).\end{align}

I found that

\begin{align}\boxed{\textbf{E1:} \quad \displaystyle \sum_{k \rm \ is \ odd} a_{n,k} =\sum_{k \rm \ is\ even} a_{n,k} \ \text{for}\ n>1\ \ \ },\end{align}

(i.e. $a_{n,1}+a_{n,3}+a_{n,5}+...=a_{n,2}+a_{n,4}+a_{n,6}+...$)


\begin{align}\boxed{\textbf{E2:}\qquad \qquad \qquad a_{n,n}=(n-1)! \qquad \quad \ }\end{align}
related to the factorial.

E1 and E2 are the two equalities I have discoverd. Moreover, \begin{align}\boxed{ \ a_{n,k}\ \text{is divisible by}\ (n-1) \text{.}\qquad \text{(i.e.}\ \ (n-1)|a_{n.k}\ \text{)}\ }\end{align}

Someone has mentioned the generalized binomial theorem which can reduce $\displaystyle \frac{d^ny}{dx^n}$ to

\begin{align} \displaystyle \frac{d^ny}{dx^n}=\sum_{k=1}^{\infty} \binom{1/n}{k} \frac{(kn)!}{\left(\left(k-1\right) n\right) !} x^{(k-1)n}\end{align}

by rewriting $y=\left(1-x^n\right)^{1/n}$ for $|x|<1$.
It could be the answer, but now I'm more interested in finding the closed-form for $a_{n,k}$.

Is there a closed-form for $\displaystyle\frac{d^ny}{dx^n}$ (in terms of $a_{n,k}$)?


Is there a closed-form for $a_{n,k}$?


  • $\begingroup$ How do you deal with non-uniqeness of representation of $y^{(n)}$ in terms of $x$ and $y$? After all, as you already said, we can express $y(x)$ in an explicit way and then play from there. $\endgroup$ – TZakrevskiy Dec 2 '16 at 16:22

We derive a recurrence formula for $a_{n,k}$ which could be helpful to find a closed formula. We start with some introductory remarks.

We can write the function $y=y(x)$ in the form \begin{align*} y(x)&=\left(1-x^n\right)^{\frac{1}{n}}\\ \end{align*} and the first derivatives are \begin{align*} \frac{d}{dx}y(x)&=-x^n\left(1-x^n\right)^{\frac{1}{n}-1}\\ \frac{d^2}{dx^2}y(x)&=-\left[(n-1)x^{2n-2}\left(1-x^n\right)^{\frac{1}{n}-2}+(n-1)x^{n-2}\left(1-x^n\right)^{\frac{1}{n}-1}\right]\\ \frac{d^3}{dx^3}y(x)&=-\left[(2n-1)(n-1)x^{3n-3}\left(1-x^n\right)^{\frac{1}{n}-3} +3(n-1)^2x^{2n-3}\left(1-x^n\right)^{\frac{1}{n}-2}\right.\\ &\qquad\qquad\left.+(n-1)(n-2)x^{n-3}\left(1-x^n\right)^{\frac{1}{n}-1}\right]\tag{1} \end{align*}

OPs example 2 has the following representation \begin{align*} \frac{d^3y}{dx^3}&=-\left(\frac{10x^6}{y^8}+\frac{12x^3}{y^5}+\frac{2}{y^2}\right)\\ &=-\left(10x^6\left(1-x^3\right)^{\frac{1}{3}-3}+12x^3\left(1-x^3\right)^{\frac{1}{3}-2} +2\left(1-x^3\right)^{\frac{1}{3}-1}\right)\tag{2} \end{align*}

The expression (2) corresponds to (1) when setting $n=3$. From (1) we derive a general formula.

Claim: We consider the $k$-th derivative of $y$ in the form \begin{align*} \frac{d^k}{dy^k}y(x)=-\sum_{j=1}^k a_{k,k-j+1}x^{jn-k}(1-x^n)^{\frac{1}{n}-j}\qquad\qquad 1\leq k\leq n\tag{3} \end{align*} with $a_{k,j}$ polynomials in $n$.

The following is valid for $1\leq j\leq k\leq n$: \begin{align*} a_{1,1}&=1\\ a_{k,k-j+1}&=((j-1)n-1)a_{k-1,k-j+1}+(jn-k+1)a_{k-1,k-j}\\ \end{align*}

We also set $a_{k,0}=a_{k,k+l}=0$ for $k,l\geq 1$.

$$ $$

The start value $a_{1,1}=\color{blue}{1}$ follows from the first derivative already stated above: \begin{align*} \frac{d}{dx}y(x)&=-\color{blue}{1}x^n\left(1-x^n\right)^{\frac{1}{n}-1}\\ \end{align*}

We obtain from (3) for $1\leq k\leq n$

\begin{align*} \frac{d^k}{dy^k}y(x)&=\frac{d}{dy}\left(\frac{d^{k-1}}{dy^{k-1}}y(x)\right)\\ &=\frac{d}{dx}\left(-\sum_{j=1}^{k-1} a_{k-1,k-j}x^{jn-k+1}(1-x^n)^{\frac{1}{n}-j}\right)\tag{4}\\ &=-\sum_{j=1}^{k-1} a_{k-1,k-j}(jn-k+1)x^{jn-k}(1-x^n)^{\frac{1}{n}-j}\\ &\qquad -\sum_{j=1}^{k-1} a_{k-1,k-j}x^{jn-k+1}\left(\frac{1}{n}-j\right)(1-x^n)^{\frac{1}{n}-j-1}\left(-nx^{n-1}\right)\tag{5}\\ &=-\sum_{j=1}^{k-1} a_{k-1,k-j}(jn-k+1)x^{jn-k}(1-x^n)^{\frac{1}{n}-j}\\ &\qquad -\sum_{j=1}^k a_{k-1,k-j}\left(jn-1\right)x^{(j+1)n-k}(1-x^n)^{\frac{1}{n}-j-1}\tag{6}\\ &=-\sum_{j=1}^k a_{k-1,k-j}(jn-k)x^{jn-k-1}(1-x^n)^{\frac{1}{n}-j}\\ &\qquad -\sum_{j=2}^{k+1} a_{k-1,k-j+1}\left((j-1)n-1\right)x^{jn-k}(1-x^n)^{\frac{1}{n}-j}\tag{7}\\ &=-\sum_{j=1}^k \left[(jn-k)a_{k-1,k-j}+\left((j-1)n-1\right)a_{k-1,k-j+1}\right] x^{jn-k}(1-x^n)^{\frac{1}{n}-j}\tag{8} \end{align*} and the claim follows.


  • In (4) we use the expression (3) and replace $k$ with $k-1$ to represent the $(k-1)$-st derivative of $y(x)$.

  • In (5) we apply the product rule to the series.

  • In (6) we collect terms of the right-hand series.

  • In (7) we shift the index $j$ of the second series by one to prepare merging of both parts.

  • In (8) we collect both series and note that for $j=1$ and $j=k$ the summands $a_{k-1,0}=a_{k-1,k}=0$.

Example: $k=1,\ldots,5$

We use the recurrence relation to show polynomials $a_{k,j}$ for small $k$ revealing thereby a regular structure.

\begin{array}{rllll} a_{1,1}&=1\\ \hline a_{2,1}&=(n-1)a_{1,1}\\ a_{2,2}&=(n-1)a_{1,1}\\ \hline a_{3,1}&=(2n-1)a_{2,1}\\ a_{3,2}&=(2n-2)a_{2,1}&+(n-1)a_{2,2}\\ a_{3,3}&=&+(n-2)a_{2,2}\\ \hline a_{4,1}&=(3n-1)a_{3,1}\\ a_{4,2}&=(3n-3)a_{3,1}&+(2n-1)a_{3,2}\\ a_{4,3}&=&+(2n-3)a_{3,2}&+(n-1)a_{3,3}\\ a_{4,4}&=&&+(n-3)a_{3,3}\\ \hline a_{5,1}&=(4n-1)a_{4,1}\\ a_{5,2}&=(4n-4)a_{4,1}&+(3n-1)a_{4,2}\\ a_{5,3}&=&+(3n-4)a_{4,2}&+(2n-1)a_{4,3}\\ a_{5,4}&=&&+(2n-4)a_{4,3}&+(n-1)a_{4,4}\\ a_{5,5}&=&&&+(n-4)a_{4,4}\\ \end{array}

  • $\begingroup$ This clearly shows the steps finding the recurrence formula! Thank you for this wonderful solution! $\endgroup$ – Tianlalu Dec 9 '16 at 12:56
  • $\begingroup$ @WongAustin: You're welcome. Many thanks for your nice comment! :-) $\endgroup$ – Markus Scheuer Dec 9 '16 at 13:06

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