# A general formula to generate functions of power series

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

$$\dfrac{1}{1+x}=1-x+x^2-x^3+x^4-x^5...$$

$$\dfrac{1}{(1+x)^2}=1-2x+3x^2-4x^3+5x^4-6x^5+7x^6...$$

I found out by accident that:

$$\dfrac{1-x}{(1+x)^3}=1-2^2x+3^2x^2-4^2x^3+5^2x^4+6^2x^5...$$

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.

• Hint: the operator $x\frac d{dx}$ changes the term $a_kx^k$ into $ka_kx^k$. – amd Nov 4 '19 at 21:58
• There were several more besides Euler and Ramanujan (as wonderful as they were). – Wlod AA Nov 5 '19 at 3:47
• The second series above is a snap. (Perhaps a few that follow take some sweat but are not too hard?). – Wlod AA Nov 5 '19 at 4:09
• 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. – Gottfried Helms Nov 18 '19 at 14:40
• @GottfriedHelms Do you have that formula, can you post it up here? – 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".)

• +1. Now, it'd be nice to add a juicy tasteful illustration, a specific series. – Wlod AA Nov 5 '19 at 11:52
• @WlodAA: Alright, I've added a nice juicy tasty illustration. =) – user21820 Nov 5 '19 at 13:09
• @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? – James Warthington Nov 5 '19 at 17:13
• Very nice! ($+\infty$ :) ). – Wlod AA Nov 5 '19 at 23:40
• @JamesWarthington: I said "as shown in the linked post"... I hope you realized that it was already explained there. – 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 $$n=k$$.

Comment:

• 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 $$k$$.

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 $$n\to \infty$$.

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