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$\displaystyle\prod_{k=1}^n(1+kx)=\underbrace{\displaystyle\sum_{k=0}^n a_k x^k}_{\text{I assumed this,it don't have to be like this}}$

I'm investigating what this means, how we can analyse this and get generalized formula.

In fact ,I thought $n-$degree equaliton's formulas.

For instance ,assume this $\displaystyle\prod_{k=1}^n(1+kx)=a_0+a_1x+....+a_{n-1}x^{n-1}+a_nx^n$

And I think we know $\displaystyle\sum \left(\dfrac{-1}{k_i}\right)=-\dfrac{a_{n-1}}{a_n}$

It's like , when ($ ax^2+bx+c=0 $) , $x_1+x_2=\dfrac{-b}{a}$

And I kept doing this , but this was gonna last to eternity...

Micheal Brain Hurts
  • 2,380

3 Answers3

4

These coefficients are essentially the unsigned Stirling numbers of the first kind $\Big[ \frac{n}{k} \Big]$: to be precise, the Stirling numbers are defined by $$x (x+1) ... (x+n-1) = \sum_{k=0}^n \Big[ \frac{n}{k} \Big] x^k.$$ The connection is that if we define $$f(x) := (1+x)(1+2x)...(1+nx),$$ then reversing the coefficients gives $$x^n f(1/x) = (x+1)(x+2)...(x+n).$$

2

We start with $$ \prod_{k = 1}^n (1 + kx) = x^n \prod_{k = 1}^n \left(\frac{1}{x} + k\right) = \sum_{j = 0}^n a_j x^j.$$ Let $y = -\frac{1}{x}$, so that this becomes $$ \frac{1}{y^n} \prod_{k = 1}^n (y - k) = \sum_{j = 0}^n a_j (-y)^{-j} = \frac{1}{y^n} \sum_{j = 0}^n (-1)^j a_j \;y^{n-j}.$$ So if we understand $$ \prod_{k = 1}^n (y-k) = \sum_{j = 0}^n (-1)^j a_j \; y^{n-j},$$ then we also understand the original product. The left hand side is very understandable, as the coefficient of $y^\ell$ of the product will be $$ \sum_{1 \leq k_1 < k_2 < \cdots < k_\ell \leq n} (-1)^\ell k_1 k_2 \cdots k_\ell.$$ Many of these are hard to understand. But the coefficient of $y^0$ is exactly $n!$, and the coefficient of $y^n$ is exactly $(-1)^n$.

These are very closely related to Vieta's formulas, Stirling numbers, and many other combinatorial questions.

davidlowryduda
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1

Notice that

$$\prod_{k=1}^n(1+kx)=\prod_{k=0}^n(1+kx)=1\times(1+x)\times(1+2x)\times\dots\times(1+nx)\\=(1+nx)\overbrace{!!\dots!!}^x\text{ or }(1+nx)!^{(x)}$$

where the long string of exclamation marks is the multifactorial, the extension of the double factorial. The Wikipedia of the factorial also has a small section on this.

As the Wikipedia states, this can be rewritten as follows:

$$\prod_{k=1}^n(1+kx)=x^n\frac{\Gamma(n+1+\frac1x)}{\Gamma(1+\frac1x)}$$

where $\Gamma$ is the Gamma function. This form allows extension to any $n,x\in\mathbb C$.

Simply Beautiful Art
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