5

Let $$ a_n = \sum_{k=0}^n \binom{n + 1}{k}b_k. $$ I am trying to write $b_n$ in terms of $a_k$.

Of course, if the binomial coefficient was $\binom{n}{k}$ instead of $\binom{n + 1}{k}$, then this is simple to do using the binomial transform, $$ c_n = \sum_{k=0}^n\binom{n}{k}d_k \iff d_n = \sum_{k=0}^n\binom{n}{k}(-1)^{n-k}c_k. $$ But in my case I'm at a loss. I tried using Pascal's identity, thinking I might be able to use the formula for the binomial transform if I could get $a_n$ expressed in terms of $b_k$ in the right way. This got me to $$ a_n = \sum_{k=0}^n \binom{n}{k}b_k + \sum_{k=0}^{n-1}\binom{n}{k}b_{k + 1}, $$ which is almost right, but not quite. Is there a simple way to do this?

mfox
  • 93
  • Define $a_k$ for $k\ne n$. – herb steinberg May 26 '21 at 21:22
  • By "write $b_n$ in terms of $a_k$" I mean "write $b_n$ as a linear combination of $a_k$, where $k = 1, \dots, n$." Here the definition of $a_k$ for $k \ne n$ is analogous to the definition of $c_k$ for $k \ne n$. – mfox May 26 '21 at 21:27
  • How is $c_k$ defined? Try to be explicit. – herb steinberg May 26 '21 at 21:36
  • Do you know generating functions? How do you show the binomial transformation? – Phicar May 26 '21 at 21:46
  • @herbsteinberg ${c_i}$ and ${d_i}$ are finite sequences related by the binomial transform. Defining one of these sequences defines the other. So, given ${d_i}$, $c_k = \sum_{i=0}^k \binom{k}{i}d_i$ for all $k = 0, \dots, n$. – mfox May 26 '21 at 21:55
  • @Phicar Not in this context, no. It's not too hard to derive the binomial transform using elementary methods, but maybe the generating function approach is better here? – mfox May 26 '21 at 21:57

2 Answers2

3

What you seek is not quite a binomial transform. Use the elementary ID $$ \binom{n+1}{k} = \frac{n+1}{n+1-k} \binom{n}{k}.$$ Define $A_n=a_n/(n+1)$ and the sum you want to invert is $$ A_n = \sum_{k=0}^n \binom{n}{k} \frac{b_k}{n-k+1} = \sum_{k=0}^n \binom{n}{k} \frac{b_{n-k}}{k+1} $$ where in the last step the sum has been performed in opposite order. Now Riordan, 'Combinatorial Identities,' eq 3.4.(17) states the inversion $$ A_n = \sum_{k=0}^n \binom{n}{k} \frac{b_{n-k}}{k+1} \implies b_n = \sum_{k=0}^n \binom{n}{k} B_k A_{n-k} $$ where $B_n$ are the Bernoulli numbers $B_0=1, \ B_1=-1/2, \ B_{2n+1} = 0.$

Example: $a_n=(2^n-1)(n+1),$ $A_n=2^{n}-1,$ then $$ b_n=\sum_{k=0}^n \binom{n}{k} B_k (2^{n-k} -1 ) = n .$$

user321120
  • 6,895
  • Thank you for your answer. I have a follow-up question: Fix $n > 0$ and define the finite sequence ${c_0, \dots, c_n}$ by $c_j = b_j (n + 1) / (n - j + 1)$. Then $a_n = \sum_{k = 0}^n \binom{n}{k}c_k$, which is (presumably) amenable to the binomial transform: $c_n = \sum_{k = 0}^n \binom{n}{k}(-1)^{n-k}a_k$. But $c_n = (n+1)b_n$. So, $b_n = \frac{1}{n+1}\sum_{k = 0}^n \binom{n}{k}(-1)^{n-k}a_k$. This doesn't seem to work. What went wrong? – mfox May 28 '21 at 15:52
  • 1
    @mfox Hello. The sequence $c_j$ must depend only on $j,$ that is, it must be independent of $n.$ – user321120 May 28 '21 at 18:36
2

Given $$ a_n = \sum_{k=0}^n \binom{n + 1}{k}b_k. \tag{1} $$ Let the exponential generating functions (e.g.f.) for $\,a_n\,$ and $\,b_n\,$ be $$ A(x) := \sum_{n=0}^\infty a_n x^n/n! \quad \text{ and } \quad B(x) := \sum_{n=0}^\infty b_n x^n/n!. \tag{2} $$ Multiply both sides of equation $(1)$ by $\,\dfrac{x^{n+1}}{(n+1)!}\,$ and sum over $\,n\ge0\,$ to get $$ \int_0^x A(t)\,dt = (e^x-1) B(x). \tag{3} $$ Thus $$ B(x) = \frac{x}{e^x-1} \left(\frac1x \int_0^x A(t)\,dt\right) \tag{4} $$ where $\,\dfrac{x}{e^x-1}\,$ is the e.g.f. of the Bernoulli numbers. The 2nd factor is the e.g.f. of $\,\dfrac{a_n}{n+1}.\,$ Thus, $$ b_n = \sum_{k=0}^n \binom{n}{k} B_{n-k} \frac{a_k}{k+1}. \tag{5} $$

Note: For the binomial transform, $\,A(x) = e^x B(x)\,$ is equivalent to $\, B(x) = e^{-x}A(x).\,$

Somos
  • 37,457
  • 3
  • 35
  • 85