Using Stirling numbers to find integers $a_k$ such that $4x^4-2x^2+5x-1=\sum_{k\ge 0}a_k(x)_k$?

Question

Find integers $a_k$ such that $4x^4-2x^2+5x-1=\sum_{k\ge 0}a_k(x)_k$

I'm guessing the Stirling numbers of the 2nd kind will be used: $\sum_{k\ge 0}S(n,k)x^k=(x)_n$

The first thing I did was I set up $$4\bigg(S(4,0)(x)_0+S(4,1)(x)_1+S(4,2)(x)_2+S(4,3)(x)_3+S(4,4)(x)_4\bigg)-2\bigg(S(2,0)(x)_0+S(2,1)(x)_1+S(2,2)(x)_2\bigg)+5\bigg(S(1,0)(x)_0+S(1,1)(x)_1\bigg)$$

Which equals $$-1(x)_0+7(x)_1+26(x)_2+24(x)_3+4(x)_4=\sum_{k=0}^{4?}?(x)_k$$

My question is simply how do I find $a_k$ and does expanding $4x^4-2x^2+5x-1$ help?

Answer 1

There are 2 kinds of Stirling numbers-those expressing a power as a sum of falling factorials and those expressing a falling factorial as a sum of powers. You could (i) find a table of Stirling numbers of the appropriate kind or (ii) generate such a table by the approprite recursive relation or (iii) find each needed power on an ad hoc basis. I;ll lead you through (iii), using the notation $$ x^{(0)}=1,x^{(1)}=x,x^{(n)}=x(x-1)...(x-n+1)$$ Thus $$x^2=x(x-1)+x=x^{(2)}+x^{(1)}$$ $$x^3=x^2x=x^{(2)}x+x^{(1)}x=x^{(2)}(x-2+2)+x^{(2)}+x^{(1)}$$ $$=x^{(3)}+2x^{(2)}+x^{(2)}+x^{(1)}=x^{(3)}+3x^{(2)}+x^{(1)}$$ $$x^4=x^{(4)}+3x^{(3)}+3(x^{(3)}+2x^{(2)})+x^{(2}+x^{(1)}$$ which I'll let you simplify.Then we'll use the forward difference notation $\Delta f(x)=f(x+1)-f(x)$ and the 'telescoping series' formula$$\sum_{x-a}^b\Delta f(x)=f(b+1)-f(a).$$ Then use the fact that $$x^{(n)}=\frac{\Delta x^{(n+1)}}{n+1}$$ to do the summation.

Answer 2

Since Stirling numbers are sought to determine $a_k$ it seems we should use ${{n}\brace {k}}$ the Stirling number of the second kind which fulfill the relation \begin{align*} \color{blue}{x^n=\sum_{k=0}^n{{n}\brace{k}} (x)_k} \end{align*}

We obtain \begin{align*} &4x^4-2x^2+5x-1=\sum_{k\geq 0} a_k(x)_k\\ &\qquad=4\sum_{k=0}^{4}{{4}\brace{k}}(x)_k-2\sum_{k=0}^2{{2}\brace{k}}(x)_k +5\sum_{k=0}^1{{1}\brace{k}}(x)_k-1\\ &\qquad=4{{4}\brace{4}}(x)_4+4{{4}\brace{3}}(x)_3+\left(4{{4}\brace{2}}-2{{2}\brace{2}}\right)(x)_{2}\\ &\qquad\quad+\left(4{{4}\brace{1}}-2{{2}\brace{1}}+5{{1}\brace{1}}\right)(x)_{1}\\ &\qquad\quad+\left(4{{4}\brace{0}}-2{{2}\brace{0}}+5{{1}\brace{0}}-1\right)(x)_{0}\\ \end{align*} with $a_k$ explicitly stated in terms of Stirling numbers of the second kind.