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:

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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.

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  • $\begingroup$ For question 1, I haven't mastered induction technique, can you do this so I can learn? $\endgroup$ – James Warthington Nov 15 '19 at 1:04
  • $\begingroup$ $\binom{n+1}{1}=\dfrac{n(n+1)}{1!}$?, I haven't mastered the binomial notation yet so I am sorry to ask. $\endgroup$ – James Warthington Nov 15 '19 at 1:06
  • $\begingroup$ So $\binom{n+2}{2}=\dfrac{n(n+1)(n+2)}{3!}$? $\endgroup$ – James Warthington Nov 15 '19 at 1:19
  • $\begingroup$ Do you mind if we go to chat section? $\endgroup$ – James Warthington Nov 15 '19 at 1:19
  • $\begingroup$ Not quite \begin{eqnarray*} \binom{n+2}{3} = \frac{n(n+1)(n+2)}{6}. \end{eqnarray*} The number at the bottom will tell you how many factors to expect at the top. $\endgroup$ – Donald Splutterwit Nov 15 '19 at 1:20

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