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Consider $$f(z) = \frac{(p_1z^2+p_2z+p_3)^n}{1-z}\quad z\in \mathbb{R}$$ where $p_1+p_2+p_3 = 1$.

I want to find a closed form solution for $f^{(k)}(0)$ where $1\le k \le 2n$.

Substituting $1-z=-t$ reduces the function to $$g(t) = -\frac{(p_1t^2+p_4t+1)^n}{t}$$ where $p_4$ is another constant. Am I correct in assuming that the $k^{th}$ derivative of $g(t)$ at $t=1$ is the same as $f^{(k)}(0)$?

Is this the correct way to proceed? I have tried the Leibniz rule and multinomial expansion but have been stuck in both approaches.

Any help would be greatly appreciated. If an exact formula is too complicated, approximations would be equally helpful.

dexter04
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1 Answers1

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I will write $a,b,c$ instead of $p_1,p_2,p_3$. Observe the equality $\frac1{1-z} = \sum z^n$ and the trinomial expansion identity, $$ (a+b+c)^n = \sum_{\substack{i,j,k\\i+j+k=n}} \binom{n}{i,j,k}a^ib^jc^k,$$ and also the formula for the coefficients of a product of two series, $$ (\sum a_N)( \sum b_N) = \sum c_N, \quad c_N = \sum_{i+j=N}a_i b_j.$$

Plugging in, we find the following horrendous equality,

$$(az^2+bz + c)^n = \sum_{i+j+k=n} \binom{n}{i,j,k} a^i b^j c^k z^{2i + j}= \sum_{N=0}^{2n} \left(\sum_{\substack{i,j,k\\ i+j+k=n\\2i+j=N}} \binom{n}{i,j,k}a^ib^jc^{k}\right)z^N =: \sum_{N=0}^\infty a_N z^N,$$ where $a_N:= 0$ for $N>2n$. This can probably be made neater but I'd call it closed form. Now use the product formula, $$f(z) = (\sum_N z^N)( \sum_N a_N z^N) = \sum_N c_N z^N,\quad c_N = \sum_{m=0}^Na_m.$$ The numbers $N! c_N = f^{(N)}(0)$ are precisely what you are looking for.

As a sanity check I will compute $c_0$ explicitly using this formula. Observe $c_0 = \sum_{N=0}^0 a_N = a_0$, and $N=0$, $2i+j=N$ means that $i=j=0$, so that $k=n$. Also $\binom{n}{0,0,n} = 1$. Hence, $$ c_0 = c^n. $$ Comparing with the original formula, $c^n$ is indeed the $y$-intercept of $f$, which is encouraging.

I'm not sure if I can simplify the above expression for $a_N$ but there's at least a more explicit formula. observe that there are 2 restrictions in the summation; therefore for fixed $k$ that we can solve for $i,j$ by solving the matrix equation $$\begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix} \binom{i}{j} = \binom{n-k}{N}.$$ So if we introduce shorthand for the $\mathfrak B$ig term, $\mathfrak B(i,j,k):=\binom{n}{i,j,k}a^ib^jc^{k}$, then $$a_N = \!\!\!\!\!\!\sum_{k=\max(0,{n-N})}^{\min(n,2n-N)} \!\!\!\!\!\!\!\mathfrak B(-n+k+N,2n-2k-N,k),$$ and hence $$f^{(M)}(0) = M! \sum_{N=0}^M \sum_{k=\max(0,{n-N})}^{\min(n,2n-N)} \!\!\!\!\!\!\! \mathfrak B(-n+k+N,2n-2k-N,k). $$ Or, if you prefer, $$f^{(M)}(0) = M! \sum_{N=0}^M \sum_{k=\max(0,{n-N})}^{\min(n,2n-N)} \binom{n}{-n+k+N,2n-2k-N,k}a^{-n+k+N}b^{2n-2k-N}c^{k}. $$

Calvin Khor
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  • Thanks for the answer. I was getting stuck at the same horrendous closed form. Any ideas on simplification? – dexter04 Mar 28 '18 at 20:23
  • For fixed $k$, the matrix equation $$ \begin{bmatrix} 1 & 1 \ 2 & 1 \end{bmatrix} \binom{i}{j} = \binom{n-k}{N} $$ is uniquely solvable for any $n,N$, i.e. there's only one value of $i,j$, explicitly $$a_N = \sum_{k=0}^n [\text{stuff at } i=-n+k+N,\ j = 2n-2k - N] $$. This is at least explicit. I can't think of anything to do with $a+b+c=1$, i guess its cool that $c_N = 1$ for $N>2n$ but thats not useful – Calvin Khor Mar 28 '18 at 20:43
  • @dexter04 added this to the body of the answer. – Calvin Khor Mar 28 '18 at 21:17