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I would like to know more about this formula (via the Worldwide Center of Mathematics website):

$$\frac{(x+y)^n}{x}=\sum_{k=0}^n\binom{n}{k}(x-ak)^{k-1}(y+ak)^{n-k}$$

What is $a$? Does the formula have a name?

Also, I would like some references, if possible. Thanks!

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    $\begingroup$ The point here is that the sum does not depend on $a$, which is surprising. $\endgroup$
    – lhf
    Jun 11, 2018 at 21:31
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    $\begingroup$ I don't know the formula name but I believe that a is a 'factor' of the polynomial. $\endgroup$
    – poetasis
    Jun 11, 2018 at 21:34

1 Answer 1

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This identity is due to Abel, and was first presented in [1]. There is more discussion in [2], available online here. That article refers to this formula as an "Abel-type generalization of the binomial theorem," but Abel had several similar generalizations in his first paper. See the Wolfram MathWorld Article for more. Wikipedia even refers to a different generalization as "Abel's Binomial Theorem.".

I'll give a quick summary. Letting $\frac{d}{dx}$ be the the derivative operator, the polynomials $x^n$ play nicely with respect to $\frac{d}{dx}$ because $\frac{d}{dx}[x^n]=nx^{n-1}$ is a polynomial of lesser degree. This allows you to express an arbitrary polynomial in terms of its $x^k$ via its Taylor series, $$f(x)=\sum_{k=0}^n \frac{((\frac{d}{dx})^kf)(0)}{k!} x^k.$$

By letting $f(x)=(x+y)^n$, we get a direct proof of the binomial theorem.

Now, instead of $x^n$, consider the series of polynomials given by $$ A_n(x,z)=x(x+nz)^{n-1} $$ Note that $A_n(x,z)$ has degree $n$ with respect to $x$, and $A_n(x,z)$ plays nicely with differentiation in the sense that $$ \frac{d}{dx}A_n(x,z)=nA_{n-1}(x+z,z) $$ so again, the derivative has a lesser degree. This allows to prove a variant to write any polynomial in terms of $A_k(x,z)$: $$ f(x)=\sum_{k=0}^n \frac{((\frac{d}{dx})^kf)(-kz)}{k!} A_k(x,z) $$ If you substitute the $f(x)=(x+y)^n$, and $z=a$, then the theorem you asked about falls out.


This phenomenon is much more general. For example, let $\Delta_1$ be the finite difference operator $\Delta_1(f)=f(x+1)-f(x)$. Then the sequence of polynomials $p_n(x)=x(x-1)(x-2)\cdots x(x-n+1)$ plays nicely with $\Delta_1$, because again we have $\Delta_1(p_n(x))=np_{n-1}(x)$ is a polynomial of lesser degree. The same logic implies that any polynomial can be written in terms of the polynomials $p_k(x)$ as $$ f(x)=\sum_{k=0}^n \frac{(\Delta_1^k\,f)(0)}{k!} p_k(x) $$ Furthermore, letting $f(x)=p_n(x+y)$, you get the surprising identity $$ p_n(x+y)=\sum_{k=0}^n \binom{n}k p_k(x)p_{n-k}(y)\tag1\label1 $$ Sequences of polynomials which satisfy $\eqref1$ are known as sequences of binomial type, and their theory is well understood, see their Wikipedia page.

[1]: N.H.Abel, Beweis eines Ausdruckes, von welchem die binomial-Formel ein einzelner Fall ist, J.reine angew.Math.1,pp.159-160(1826)

[2]: Kuriyama, K., & Furuichi, S. (2014). GENERALIZED DIFFERENCES AND ABEL’S TYPE BINOMIAL THEOREMS. International Journal of Pure and Applied Mathematics, 96(2). https://doi.org/10.12732/ijpam.v96i2.3

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