How is the binomial expansion of the vectors?

Question

I'm trying to find out if there is an attempt to define binomial expansion of vectors. i.e

$$(\overrightarrow a + \overrightarrow b)^n = ?? $$

I tried to google around this (e.g : binomial expansion of vectors), but simple searches do not give any useful keywords as how this approach is named.

I'm trying to visualize how the following generalization goes. When two vectors are on the number line (one dimension), their sum is just the magnitude addition and the result is simply lying in the same number line. When the lines lie in general two dimensional space their 'sum' is equivalent to the square of the magnitude of their vector sum (which is essentially the binomial expansion of $(a + b)^2$ ).

So how is the interpretation of $ (a+b)^3 $? Definitely it is not related to the 'normal' vector sum of the two vectors in 3d (that would be simply $(a + b)^2$).

Answer 1

In order to calculate the expansion, you need to be able to do both addition and multiplication, and both of these operations need to return another "vector". The fancy mathematical way to say this is that the objects $a$ and $b$ need to belong to a ring (which is just a structure that has addition and multiplication operations).

In order for the formula to be true (as opposed to merely making sense), you need to be in an algebraic structure that has commutative addition and multiplication operations (like a commutative ring). Without commutativity, even the $n=2$ formula is not true -- you get $(a+b)^2 = a^2 + ab+ba+b^2$, and the right-hand side isn't necessarily equal to $a^2 + 2ab+b^2$.

Vectors (in the usual sense of the word) don't have a commutative (or even associative) multiplication operation, so the expansion doesn't even make sense. You can interpret $(a+b)^2$ as a dot product: $(a+b)\cdot(a+b)$. This works, but it gives you a scalar. So then what is $(a+b)^3$? You could say that it's the product of the scalar $(a+b)^2$ and the vector $(a+b)$, but this process doesn't seem to lead to anything useful.

See also this question.

Answer 2

If we use Dot product then $(a+b)\cdot(a+b)=a\cdot a+2a\cdot b+b\cdot b$.

Since we can't define $a^3=(a\cdot a)\cdot a$ we can't generalize this for $n\geq 3$.

The reason that we can't define $a^3$ is that $(a\cdot a)\in \mathbb R$ but $a\in \mathbb R^n$ and Dot product is defined only between two elements in the same dimension.

Answer 3

Here's a possible use of $(a+b)^n$ for vectors $a$ and $b$ and any integer $n$ (although it doesn't agree with the analogy for $n=2$):

Think of vectors as homogeneous polynomials of degree 1 on the dual of the linear space. Then $(a+b)^n$ is a homogeneous polynomial of degree $n$. For example, if $a=(1,2)$ and $y=(2,3)$ then one can think of $a$ as $x+2y$ and of $b$ as $2x+3y$, hence $(a+b)^n = (3x+5y)^n$.

Answer 4

It's "weird", but you could encode your vectors down the diagonals of matrices. Then you have embedded your problem into an algebra (a ring with commutative multiplication -- although matrix multiplication is not generally commutative, the multiplication of these diagonal matrices is). As with some of the other answers, I'm not sure that this leads to anything interesting, since it's pretty much just using the Hadamard product.

\begin{align*} (\vec{a} + \vec{b})^3 &= \left( \begin{pmatrix} a_1 & & \Large{0} \\ & \ddots & \\ \Large{0} & & a_n \end{pmatrix} + \begin{pmatrix} b_1 & & \Large{0} \\ & \ddots & \\ \Large{0} & & b_n \end{pmatrix}\right)^3 \\ &= \overrightarrow{a^3} + 3\overrightarrow{a^2 b} + 3 \overrightarrow{a b^2} + \overrightarrow{b^3} \end{align*} I seriously doubt there's any sort of interesting geometric story told by this equation.

Answer 5

You can take a binomial type expansion of the vector expression $$|a-b|^n,$$ where $|\cdot|$ is the norm. For instance, in $\mathbb{R}^2$ consider $f(x) = |x+y|^n$ Taylor expanded about the point $x = (0, 0)$. For $n=2$ we have \begin{align} f(x) & = f(0) + x_1f_{x_1}(0) + x_2f_{x_2}(0) + \frac{1}{2!}\bigg(x_1^2f_{x_1x_1}(0) + 2x_1x_2f_{x_1x_2}(0) + x_2^2f_{x_2x_2}(0)\bigg) \\ & = |y|^2 + 2\langle x,y\rangle + |x|^2. \end{align} For $n = 4$ you would obtain $$ |x+y|^4 = |y|^4 + 4 \langle x, y\rangle|y|^2 + (2|x|^2|y|^2 + 4\langle x, y \rangle^2) + 4 \langle x, y\rangle|x|^2 + |x|^4. $$