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I am reading through Concrete Mathematics (Graham, Knuth, Patashnik) and they show you can express regular powers in terms of falling powers; starting with

$$ k^2 = k^\underline{2} + k^\underline{1} $$

Then they go on to show $k^3$ and that a "simple calculation" reveals

$$ k^3 = k^\underline{3} + 3k^\underline{2} + k^\underline{1} $$

I am having trouble understand what that simple calculation is. I also found this question which was some help; to quote

It is easy to see that $n^1 = n^\underline{1}$ and $n^2 = n^\underline{2} + n^\underline{1}$. With these base cases, one can recursively express any normal power of $n^k$ as a sum of $n^\underline{k}$, yielding a $k$th order polynomial. Then, substitute falling-power-expansions for all terms of order less than $k$, expressing $n^\underline{k}$ as $n^k$ plus some lower-order falling powers. Rearrange to put $n^k$ on the LHS and all falling powers on the RHS. For example, one can find that $n^3 = n^\underline{3} + 3n^\underline{2} + n^\underline{1}$

I am trying this but obviously not understanding the process for $n^3$

$$ \begin{align} n(n-1)(n-2) &= n(n^2 - 3n + 2) \\ &= n^3 - 3n^2 + 2n \\ -n^3 &= -3n(n-1) +n -2n \\ -n^3 &= -3n(n-1) -n \\ n^3 &= 3n(n-1) + n \\ n^3 &\ne 3n^\underline{2} +n^\underline{1} \end{align} $$

Which is clearly not correct.

Yes, there is a better way with Stirling Numbers of the Second Kind but I want to understand the "naive" way first.

Friedrich 'Fred' Clausen
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1 Answers1

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You already know that $n^1=n^{\underline1}$ and $n^2=n^{\underline2}+n^{\underline 1}$. Then

$$\begin{align*} n^{\underline3}&=n(n-1)(n-2)\\ &=n^3-3n^2+2n\\ &=n^3-3\left(n^{\underline2}+n^{\underline 1}\right)+2n^{\underline1}\\ &=n^3-3n^{\underline2}-n^{\underline1}\,, \end{align*}$$

so

$$n^3=n^{\underline3}+3n^{\underline2}+n^{\underline 1}\,.$$

Brian M. Scott
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