Is there an "inverted" dot product?

The dot product of vectors $\mathbf{a}$ and $\mathbf{b}$ is defined as: $$\mathbf{a} \cdot \mathbf{b} =\sum_{i=1}^{n}a_{i}b_{i}=a_{1}b_{1}+a_{2}b_{2}+\cdots +a_{n}b_{n}$$

What about the quantity? $$\mathbf{a} \star \mathbf{b} = \prod_{i=1}^{n} (a_{i} + b_{i}) = (a_{1} +b_{1})\,(a_{2}+b_{2})\cdots \,(a_{n}+b_{n})$$

Does it have a name?

"Dot sum" seems largely inappropriate. Come to think of it, I find it interesting that the dot product is named as such, given that it is, after all, a "sum of products" (although I am aware that properties of $\mathbf{a} \cdot{} \mathbf{b}$, in particular distributivity, make it a meaningful name).

$\mathbf{a} \star \mathbf{b}$ is commutative and has the following property:

$\mathbf{a} \star (\mathbf{b} + \mathbf{c}) = \mathbf{b} \star (\mathbf{a} + \mathbf{c}) = \mathbf{c} \star (\mathbf{a} + \mathbf{b})$

Solution 1:

Too long for a comment, but I'll list some properties below, in hopes some idea comes up.

• ${\bf a}\star {\bf b}={\bf b}\star {\bf a}$;
• $(c{\bf a})\star (c {\bf b})=c^n ({\bf a}\star {\bf b})$;
• $({\bf a+b})\star {\bf c} = ({\bf a+c})\star {\bf b} = ({\bf b+c})\star {\bf a}$;
• ${\bf a}\star {\bf a} = 2^n a_1\cdots a_n$;
• ${\bf a}\star {\bf 0} = a_1\cdots a_n$;
• $(c{\bf a})\star {\bf b} = c^n ({\bf a}\star ({\bf b}/c))$;
• ${\bf a}\star (-{\bf a}) = 0$;
• ${\bf 1}\star {\bf 0} = 1$, where ${\bf 1} = (1,\ldots,1)$;
• $\sigma({\bf a}) \star \sigma({\bf b}) = {\bf a}\star {\bf b}$, where $\sigma \in S_n$ acts as $\sigma(a_1,\ldots,a_n) \doteq (a_{\sigma(1)},\ldots,a_{\sigma(n)})$.

Solution 2:

I don't know if it has a particular name, but it is essentially a peculiar type of convolution. Note that $$ \prod_{i}(a_{i} + b_{i}) = \sum_{X \subseteq [n]} \left( \prod_{i \in X} a_{i} \right) \left( \prod_{i \in X^{c}} b_{i} \right), $$ where $X^{c} = [n] \setminus X$ and $[n] = \{1, 2, \dots n\}$. In other words, if we define $f_{a}, f_{b}$ via $$ f_{a}(X) = \prod_{i \in X}a_{i}, $$ then $$ a \star b = (f_{a} \ast f_{b})([n]) $$ where $\ast$ denotes the convolution product $$ (f \ast g)(Y) = \sum_{X \subseteq Y} f(X)g(Y \setminus X). $$ To learn more about this, I would recommend reading about multiplicative functions and Moebius inversion in number theory. I don't know if there is a general theory concerning this, but the notion of convolutions comes up in many contexts (see this wikipedia article, and another on its role in number theory).

Edit: For what it's worth, the operation is not a vector operation in linear-algebraic sense. That is, it is not preserved under change-of-basis. In fact, it is not even preserved under orthogonal change-of-basis (aka rotation). For example, consider $a = (3,4) \in \mathbb{R}^{2}$. Note that $a \star a = 32$. Then we apply the proper rotation $T$ defined by $T(a) = (5, 0)$. Then we see $T(a) \star T(a) = 0$.

Solution 3:

I'd write$$ \mathbf{a} \star \mathbf{b} = \prod_{i=1}^{n} \left(a_{i} + b_{i}\right)=(a_{1} +b_{1})\,(a_{2}+b_{2})\cdots \,(a_{n}+b_{n}) $$up as$$ \mathrm{prod}\left(\mathbf{a}+\mathbf{b}\right), $$where $\mathrm{prod}\left(\right)$ is the array product function,$$ \mathrm{prod}\left(\mathbf{x}\right)~{\equiv}~\prod_{{\forall}\text{indices}~i}{x_i}. $$

As discussed in @6005's answer, the nice properties of this operation follow from the simplicity of this construction.

Alternatives: "Log-norm" or "product-norm"

There're $p$-norms,$$ {\left|\left|\mathbf{x}\right|\right|}_p~{\equiv}~{\left(\sum_{{\forall}i}{\left|{x_i}\right|}^p\right)}^{p^{-1}}, $$which is more generally defined as:

1. Some element-wise transform, e.g. ${\left(\right)}^{p}$, is applied.

2. The transformed elements are summed.

3. The reverse transform, e.g. ${\left(\right)}^{p^{-1}}$, is applied.

An array product can be written in this form where the element-wise transform is $\ln{\left(\right)}$ and its reverse is $\exp{\left(\right)}$, i.e.$$ {\left|\left|\mathbf{x}\right|\right|}_{\text{log}}~{\equiv}~\exp{\left(\sum_{{\forall}i}{\ln{{x_i}}}\right)}. $$ So, this might be written up as$$ {\left|\left|\mathbf{a}+\mathbf{b}\right|\right|}_{\text{log}}, $$which might be called the "log-norm" by analogy to $p$-norms.

This occurs in product topology, so we might call it the "product norm".

Suggestion: Try to work in the log-transformed domain

From a practical perspective, I'd try to work in the log-transformed domain when possible. I mean,$$\mathrm{prod}\left(\mathbf{a}+\mathbf{b}\right)$$is a nice, clean expression, but$$\mathbf{a}+\mathbf{b}$$is really clean.

It'd be easier to describe exactly how to do this if your problem domain were known, but an abstract gist would be like:

1. At current, you're basically adding vectors up and multiplying out the sum's elements.

2. The cleanness of this system is obstructed by its formulation as a web of additions and multiplications; either one or the other would be far more systematic.

3. You can convert one into the other using exponential/logarithmic transforms.

4. So, find your inputs/outputs, and $\exp{}$/$\ln{}$ 'em.

For precedents, this sort of technique is widely used in specific cases like Laplace transforms and Fourier transforms. So, it seems like you should basically try to create your own log-scale transform domain appropriate to the problem.

Solution 4:

Many operations are better expressed as the combination of several simpler operations. For instance, if I were to define $$ \vec{u} \oplus \vec{v} := \frac{u + v}{2}, \tag{1} $$ this may have some nice properties -- such as commutativity, $(\vec{u} \oplus \vec{v}) \oplus (\vec{w} \oplus \vec{x}) = (\vec{u} \oplus \vec{w}) \oplus (\vec{v} \oplus \vec{x})$, $\vec{u} \oplus \vec{u} = \vec{u}$, and so on.

However, in most contexts, it would be better to only define addition, and scaling a vector by $\frac12$. This concept is a derived concept and it does not need its own name. Its properties are best expressed as resulting from properties of $+$ and of scaling by $\frac12$.

Nat's answer expresses your operation, $\star$, as the composition of two simpler operations: $$ \vec{a} \star \vec{b} = \mathrm{prod}\left(\vec{a}+\vec{b}\right), $$ where $\mathrm{prod}: \mathbb{R}^n \to \mathbb{R}$ is the product of a vector's coordinates. So far, every property I have seen on this page about $\star$ is derived already from properties of $+$ and properties of prod. In particular, absent any algebraic expression that seems nicer to write in terms of $\star$, a working mathematician will always prefer to express it in terms of $+$ and $\mathrm{prod}$.

Now, it may be that you find an expression or a context where $\star$ is the right way to express things (it would have to be something very different from usual linear algebra). If you do, your operation is useful. But I think $\star$ obscures rather than illuminates so far, while using $+$ and $\mathrm{prod}$ illuminates.

Solution 5:

One obvious problem seems to be that it doesn't behave well under a change of basis. Consider $\vec{a}=\vec{b}=\left(1,0\right)$. Then $\vec a\star \vec b=0$, but after a change to another (still orthonormal) basis, we could have $\vec{a}=\vec{b}=\left(\frac{1}{\sqrt2},\frac{1}{\sqrt{2}}\right)$ and $\vec a\star \vec b=2$.

You can work out the general expression for how the "star-product" transforms, but this example already shows that a well-behaved transformation changes the result from zero to nonzero. In a $d$-dimensional space, you could e.g. freely pick $d-1$ vectors $\vec v_a$ and choose a basis that their last component is zero. Then all products $\vec v_a\star \vec v_b=0$.

Hence, the star product seems to be an operation on $d$-tuples of numbers, but not on vectors.

UPDATE

We can turn this into a proper vector expression if we include some extra structure: Assume the vector space is equipped with a scalar product and fix an orthonormal basis $\vec e_{(i)}$. In this basis, define the product as you did. Then the obvious generalisation is $$\vec a\star\vec b= \prod_i\left<\vec a+\vec b,\vec e_{(i)}\right>\,.$$ Geometrically, this measures the $d$-dimensional volume of the box (hyperrectangle) with $\vec a+\vec b$ as the 'space diagonal'. The catch of course is that now that the volume depends on the orthonormal set $\left\{\vec e_{(i)}\right\}$, so that in any other basis, the expression will look quite more complicated.

Of course, as has been mentioned by Nat and 6005, this would be more easily framed in terms of the expression on the rhs applied to the sum of $\vec a$ and $\vec b$. (You may call that $\text{prod}$, but in general it is not just a product of components.) The $\star$ operation itself does not seem to be very 'producty' as far as $\vec a$ and $\vec b$ are concerned.