In mathematics, it is common to consider functions as scalars — objects defined by a single numerical value at each point in the domain. However, the vector approach allows us to reveal the internal structure of a function and make its decomposition into "components" clear. In this note, we will show that any sufficiently smooth scalar function \(f(x)\) will naturally coincide with the scalar product of the vector analogue \(\mathbf{f}(x)\) with itself: \[ f(x)^2 = \mathbf{f}(x) \cdot \mathbf{f}(x) \] This transformation opens up new ways of analyzing functions, allowing us to take a different look at familiar mathematical constructs. Next, we will formulate and prove the corresponding theorem, and then give examples of its application.

The vector analogue of the function \(f(x)\) will look like this: \[ \mathbf{f}(x) = \mathbf{j_0} a_0 + \mathbf{j_1} a_1 x^{1/2 } + \mathbf{j_2} a_2 x^{1} + \ldots + \mathbf{j_n} a_n x^{n/2} \] in this case: \(\mathbf{j_0} \ldots \mathbf{j_n}\) - unit vectors of the orthonormal basis, which implies the following conditions: \[ \mathbf{j_n}\cdot \mathbf{j_m} = \begin{cases} 0, & \mbox{if } n \neq m \\ 1, & \mbox{if } n = m \end{cases}\] The number is an integer: \(n \in 0,1,2,3, \ldots \infty\). Then the coefficients of the unit vectors will be found as follows: \[ a_n = \left. {\pm \sqrt{[f(x)^2]^{(n)} \over n!}}\, \right|_{\large x=0} \] where \((n)\) is the order of the derivative with respect to the variable \(x\).

We will further call the operation itself transformation of a scalar function into a vector one.

As you know, any function can be expanded in a power series Maclaurin [1]. The square of any function will be the function that, in turn, also can be expanded in such a series: \[f(x)^2 = a_0^2 + a_1^2 x + a_2^2 x^2 + a_3^2 x^3 + \ldots + a_n^2 x^n \tag{2.1}\] where: \(n\) — arbitrarily big integer. Then: \(f(0)^2 = a_0^2\) where: \(a_0 = \pm f(0)\). Next, we will take the derivative of this series at \(x\), then find the value of the function is zero: \[ [f(x)^2]^{(1)} = a_1^2 + a_2 2^2 x^1 + 3 a_3^3 x^2 + \ldots + n a_n^2 x^{n-1}, \quad [f(0)^2]^{(1)} = a_1^2 \tag{2.2}\] \[ [f(x)^2]^{(2)} = 2 a_2^2 + 3\cdot2 a_3^2 x^1 + \ldots + n (n-1) a_n^2 x^{n-2}, \quad [f(0)^2]^{(2)} = 2 a_2^2 \tag{2.3}\] \[ \ldots \] \[ [f(x)^2]^{(n)} = n(n-1)(n-2)\ldots 3\cdot2 a_n^2, \quad [f(0)^2]^{(n)} = n!\, a_n^2 \tag{2.4}\] Thus, the values of the coefficients will be according to the formula: \[\tag{2.5} \left. {a_n = \pm \sqrt{[f(x)^2]^{(n)} \over n!}}\, \right|_{\large x=0}, \quad a_0 = \pm f(0) \] And the vector will be drawn from them: \[\mathbf{f}(x) = \left\{ a_0,\, a_1 x^{1/2 },\, a_2 x^{1},\, \ldots,\, a_n x^{n/2} \right\} \tag{2.6}\] Therefore performed and the condition: \(f(x)^2 = \mathbf{f}(x) \cdot \mathbf{f}(x)\). The theorem is proved.

Now it is evident that almost any scalar function (except those that have a problem with the derivative) can be represented as a vector: \[f(x) \to \mathbf{f}(x) \tag{2.7}\] And the decomposition of the vector in (2.5-2.6) may be not unique.

Example 1:  \(f(x) = 1 + x\)
\[f(x)^2 = (1 + x)^2,\, f(0)^2 = 1\] \[[f(0)^2]^{'} = [(1 + x)^2]_x^{'} = 2(1 + x) \big |_{x=0} = 2\] \[[f(0)^2]^{''} = [(1 + x)^2]_x^{''} = 2\] Therefore, the vector from scalar functions is: \[\mathbf{f}(x) = \left\{\pm 1,\, \pm \sqrt{2x}, \pm x \right\} \tag{2.8}\] Below is a geometric interpretation of such a transformation for which only one option \(\pm\) is selected:

Fig.3. An example of converting a scalar function \(f(x)=1+x\) into a vector \(\mathbf{f}(x)\)

Let us check that in this example the main condition of the theorem proved above is satisfied: \(f(x)^2 = \mathbf{f}(x)^2 = (1 + x)^2\)

 

Example 2:  \(f(x) = {1 \over \sqrt{1 - x^2}} \)
Let us denote \(x^2 = y\) then: \[f(y)^2 = {1 \over 1 - y}\, \Rightarrow\, [f(0)^2]^{(n)} = n!\, \Rightarrow\, a_n = \pm 1 \] Hence, the vector from \(y\) is: \[\mathbf{f}(y) = \left\{\pm 1,\, \pm y^{1/2},\, \pm y^{1},\, \ldots,\, \pm y^{n/2} \right\} \] and the desired vector from a scalar elementary function like this: \[\mathbf{f}(x) = \left\{\pm 1,\, \pm x,\, \pm x^2,\, \ldots,\, \pm x^n \right\} \tag{2.9}\] Other examples of transformations of some scalar periodic and hyperbolic functions are given in the following works: #1 and #2. It should be noted here that if, in the process of transforming a scalar function into a vector function, a minus one appears under the square root sign, then such a transformation is extended to the algebra of hyperbolic numbers.

The theorem proved here allows us to look at familiar space from a slightly different angle, and then we show how it can look in fact, from the point of view of mathematics :)