I have been playing with Maclaurin series lately, I have been able to come across this:



I found out by accident that:


I found in an old paper of Euler that this can continue on with these functions:

$\dfrac{1-4x+x^2}{(1+x)^4}=1-2^3x+3^3x^2-4^3x^3+5^3x^4+6^3x^5...$ $\dfrac{1-11x+11x^2-x^3}{(1+x)^5}=1-2^4x+3^4x^4-4^4x^3+5^4x^4+6^4x^5...$ $\dfrac{1-26x+66x^2-26x^3+x^4}{(1+x)^5}=1-2^5x+3^5x^4-4^5x^3+5^5x^4+6^5x^5...$ $\dfrac{1-57x+320x^2-302x^3+57x^4-x^5}{(1+x)^5}=1-2^6x+3^6x^4-4^6x^3+5^6x^4+6^6x^5...$

and so on... Is there a general formula to generate the functions on the right hand side? How did Euler calculate these series? I have to say I deeply respect him since only he and Ramanujan know how to play with series.

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    $\begingroup$ Hint: the operator $x\frac d{dx}$ changes the term $a_kx^k$ into $ka_kx^k$. $\endgroup$ – amd Nov 4 '19 at 21:58
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    $\begingroup$ There were several more besides Euler and Ramanujan (as wonderful as they were). $\endgroup$ – Wlod AA Nov 5 '19 at 3:47
  • $\begingroup$ The second series above is a snap. (Perhaps a few that follow take some sweat but are not too hard?). $\endgroup$ – Wlod AA Nov 5 '19 at 4:09
  • $\begingroup$ The coefficients in the numerators are also from the "Eulerian triangle" or "triangle of Eulerian numbers". There is an easy recursion-formula to arrive from row $r$ at row $r+1$ in your above list. Moreover, there is a formula for computing the coefficients directly from the row-index. $\endgroup$ – Gottfried Helms Nov 18 '19 at 14:40
  • $\begingroup$ @GottfriedHelms Do you have that formula, can you post it up here? $\endgroup$ – James Warthington Nov 18 '19 at 16:47

The most efficient way to obtain such formulae is to compute the Newton series for $k$-th powers, and then use the fact that $1/(1-x)^{k+1} = \sum_{n=0}^∞ \binom{n+k}{k} x^n$ for $|x| < 1$, which is easy to prove by induction (or by observing that the coefficients for the series are a column of Pascal's triangle).

For example, we can easily get $n^3 = 1 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}$, as shown in the linked post, and hence

$\sum_{n=0}^∞ (n+1)^3 x^n = \sum_{n=0}^∞ \left( 1 \binom{n+1}{1} + 6 \binom{n+1}{2} + 6 \binom{n+1}{3} \right) x^n$

    $ = \sum_{n=0}^∞ \left( 1 \binom{n+1}{1} + 6 \binom{n+2}{2} x + 6 \binom{n+3}{3} x^2 \right) x^n$   [since $\binom{n}{k} = 0$ for $0 ≤ n < k$]

    $ = 1/(1-x)^2 + 6x/(1-x)^3 + 6x^2/(1-x)^4$   [for $|x| < 1$]

    $ = \left( (1-x)^2 + 6x(1-x) + 6x^2 \right) / (1-x)^4$

    $ = \left( 1 + 4x + x^2 \right) / (1-x)^4$

(Substituting $x$ with $-x$ gives the first series cited in the question from "an old paper of Euler".)

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  • $\begingroup$ +1. Now, it'd be nice to add a juicy tasteful illustration, a specific series. $\endgroup$ – Wlod AA Nov 5 '19 at 11:52
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    $\begingroup$ @WlodAA: Alright, I've added a nice juicy tasty illustration. =) $\endgroup$ – user21820 Nov 5 '19 at 13:09
  • $\begingroup$ @user21820: I have looked into the Pascal triangle and cannot find any pair that is $(1, 6, 6)$. You mean it is the column, not the row, right? $\endgroup$ – James Warthington Nov 5 '19 at 17:13
  • $\begingroup$ Very nice! ($+\infty$ :) ). $\endgroup$ – Wlod AA Nov 5 '19 at 23:40
  • $\begingroup$ @JamesWarthington: I said "as shown in the linked post"... I hope you realized that it was already explained there. $\endgroup$ – user21820 Nov 6 '19 at 2:59

We are looking for polynomials $p_k(x)$ with \begin{align*} \frac{p_k(x)}{(1+x)^{k+1}}=\sum_{j=0}^\infty (-1)^j(j+1)^kx^j\qquad\qquad k\geq 0 \end{align*}

We can find $p_k(x)$ as follows: \begin{align*} \color{blue}{p_k(x)}&=\left(\sum_{j=0}^\infty(-1)^j(j+1)^kx^j\right)(1+x)^{k+1}\\ &=\left(\sum_{j=0}^\infty(-1)^j(j+1)^kx^j\right)\left(\sum_{l=0}^{k+1}\binom{k+1}{l}x^l\right)\\ &=\sum_{n=0}^\infty\left(\sum_{{j+l=n}\atop{j,l\geq 0}}\binom{k+1}{l}(-1)^j(j+1)^k\right)x^n\\ &\,\,\color{blue}{=\sum_{n=0}^\infty\left(\sum_{l=0}^{\min\{n,k+1\}}\binom{k+1}{l}(-1)^{n-l}(n-l+1)^k\right)x^n}\tag{1} \end{align*}

OP's result indicates that for $k\geq 1$ we expect $p_k(x)$ to be a polynomial of degree less than or equal to $k-1$. We therefore want to show that for $k\geq 1$ \begin{align*} p_k(x)=\sum_{n=0}^{\color{blue}{k-1}}\left(\sum_{l=0}^{\color{blue}{n}}\binom{k+1}{l}(-1)^{n-l}(n-l+1)^k\right)x^n\tag{2} \end{align*}

In order to show (2) it is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$. This way we can write for instance \begin{align*} n^k=k![x^k]e^{nx}\tag{3} \end{align*}

We obtain for $n\geq k+1$

\begin{align*} \color{blue}{\sum_{l\geq 0}}&\color{blue}{\binom{k+1}{l}(-1)^{n-l}(n-l+1)^k}\\ &=\sum_{l\geq 0}\binom{k+1}{l}(-1)^{n-l}k![x^k]e^{x(n-l+1)}\tag{4}\\ &=(-1)^nk![x^k]e^{x(n+1)}\sum_{l\geq 0}\binom{k+1}{l}(-1)^le^{-lx}\tag{5}\\ &=(-1)^nk![x^k]e^{x(n+1)}\left(1-e^{-x}\right)^{k+1}\tag{6}\\ &=(-1)^nk![x^k]e^{x(n-k)}\left(e^x-1\right)^{k+1}\tag{7}\\ &\,\,\color{blue}{=0}\tag{8} \end{align*} and the claim (2) follows for $n\geq k+1$. Similarly the claim can be shown for $n=k$.


  • In (4) we apply the coefficient of operator according to (3).

  • In (5) we do some rearrangements.

  • In (6) we apply the binomial theorem.

  • In (7) we factor out $e^{-x(k+1)}$.

  • In (8) we note that $(e^x-1)^{k+1}=x^{k+1}+\cdots$ consists of terms with powers of $x$ greater than $k$.

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Given any analytic function $f ,f(x)=\sum_{n=0}^{\infty}f^n(0)x^n/n!$.You are guaranteed to get a series representation iff $R_n(x)=f(x)-\sum_{k=0}^{n}f^k(0)x^k/k! \to 0$ pointwise as $n\to \infty$.

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  • $\begingroup$ Well, this is about actual formulas (very fast algorithms?) and not just about existence. $\endgroup$ – Wlod AA Nov 5 '19 at 3:53
  • $\begingroup$ I have mentioned the algorithm also in the 1st line $\endgroup$ – Kishalay Sarkar Nov 5 '19 at 3:57
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    $\begingroup$ It's about fast (efficient) algorithms. If you have them, then please answer the OP's question. $\endgroup$ – Wlod AA Nov 5 '19 at 4:02
  • $\begingroup$ I am not sure see if this helps books.google.co.in/… $\endgroup$ – Kishalay Sarkar Nov 5 '19 at 4:04
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    $\begingroup$ @WlodAA: See my answer, which gives a method faster than any method based on differentiation. $\endgroup$ – user21820 Nov 5 '19 at 11:34

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