In this topic: A general formula to generate functions of power series
I have asked for a general method to calculate the generating function of Euler's series. User21820 seemingly provided an exhaustive answer. However, there is still some points I don't understand. I have included that in this image below:
I have these following questions:
1) I don't know understand how he obtains: $(\dfrac{1}{1-x})^{k+1}=\sum_{n=0}^{\infty}\binom{n+k}{k}x^n$
2) In the first line, it seems to me that there is a typo, does he mean
$\sum_{n=0}^∞ (n+1)^3 x^n = \sum_{n=0}^∞ \left( 1 \binom{n+1}{1} + 6 \binom{n+2}{2} + 6 \binom{n+3}{3} \right) x^n$ instead of
$\sum_{n=0}^∞ (n+1)^3 x^n = \sum_{n=0}^∞ \left( 1 \binom{n+1}{1} + 6 \binom{n+1}{2} + 6 \binom{n+1}{3} \right) x^n$?
3) In the second line, where does the $x^2$ come from?
4) I don't understand how did he obtain these coefficients $(1, 6, 6)$
5) How do I obtain coefficients for higher order of $n$? For example what is the coefficients for $n=0, n=1, n=2$?
Question 1. : This is a well known formula and can easily be shown by induction.
Question 2. : No. This line is sound.
Question 3. : Did you notice the factor of $x$ in the second term ? ... The sums are being reindexed (in light of terms that are zero.)
Question 4. : Do the algebra ! \begin{eqnarray*} n^3= \alpha \frac{n(n-1)(n-2)}{6} + \beta \frac{n(n-1)}{2} + \gamma n. \end{eqnarray*}
Question 5. : These coefficients in general are can be expressed in terms of the Stirling numbers of the second kind.
\begin{eqnarray*} n^m= \sum_{k=1}^{m} k! S(n,k) \binom{n}{k}. \end{eqnarray*}
Edit :We want to show \begin{eqnarray*} \frac{1}{(1-x)^{k+1}} = \sum_{k=0}^{\infty} \binom{n+k}{k}x^k. \end{eqnarray*} The base case is the geometric sum \begin{eqnarray*} \frac{1}{(1-x)} = \sum_{k=0}^{\infty} x^k. \end{eqnarray*} Now assume the formula is true for $k$ and multiply by $\frac{1}{(1-x)}$ and use the hockey stick identity.