# In the formula $\frac{(x+y)^n}{x}=\sum_{k=0}^n\binom{n}{k}(x-ak)^{k-1}(y+ak)^{n-k}$, what is $a$? What is this formula named?

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

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

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

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

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