2019-10-03
2. The theorem on the transformation of scalar to vector
Mathematicians like to prove everything very strictly and according to its special rules. Because we will be it is important, that we allow ourselves some liberties, but "leave Caesar what is Caesar's". However, here we prove that the scalar function \(f(x)\) can be decomposed into the vector \(\mathbf{f}(x)\) such that: \(f(x)^2 = \mathbf{f}(x) \cdot \mathbf{f}(x)\).
The vector will look like this: \(\mathbf{f}(x) = \mathbf{j_0} a_0 + \mathbf{j_1} a_1 x^{1/2 } + \mathbf{j_2} a_2 x^{1} + \ldots + \mathbf{j_n} a_n x^{n/2}\),
thus: \(\mathbf{j_0} \ldots \mathbf{j_n}\) are 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 coefficients of the unit vectors will be like this: \(a_n = \pm \sqrt{[f(0)^2]^{(n)} \over n!} \), where \((n)\) — order derivative of a variable \(x\).
The vector will look like this: \(\mathbf{f}(x) = \mathbf{j_0} a_0 + \mathbf{j_1} a_1 x^{1/2 } + \mathbf{j_2} a_2 x^{1} + \ldots + \mathbf{j_n} a_n x^{n/2}\),
thus: \(\mathbf{j_0} \ldots \mathbf{j_n}\) are 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 coefficients of the unit vectors will be like this: \(a_n = \pm \sqrt{[f(0)^2]^{(n)} \over n!} \), where \((n)\) — order derivative of a variable \(x\).
More modern but less descriptive entry of the vector is: \(\mathbf{f}(x) = \left\{ a_0,\, a_1 x^{1/2 },\, a_2 x^{1},\, \ldots,\, a_n x^{n/2} \right\}\). In the further presentation we will use old and new forms.
Proof
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: \[a_n = \pm \sqrt{[f(0)^2]^{(n)} \over n!}, \quad a_0 = \pm f(0) \tag{2.5}\] 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.
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: \[a_n = \pm \sqrt{[f(0)^2]^{(n)} \over n!}, \quad a_0 = \pm f(0) \tag{2.5}\] 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.
Examples of transformations
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: \[\mathbf{f}(x) = \left\{\pm 1,\, \pm \sqrt{2x}, \pm x \right\} \tag{2.8}\] The following is a geometric interpretation of such a transformation:
\[f(x)^2 = (1 + x)^2,\, f(0)^2 = 1\] \[[f(0)^2]^{'} = [(1 + x)^2]_x^{'} = 2(1 + x) \big |_{x=0} = 2\] \[[f(0)^2]^{''} = [(1 + x)^2]_x^{''} = 2\] Therefore, the vector from scalar functions is: \[\mathbf{f}(x) = \left\{\pm 1,\, \pm \sqrt{2x}, \pm x \right\} \tag{2.8}\] The following is a geometric interpretation of such a transformation:
Fig.3. An example of converting a scalar function \(f(x)=1+x\) into a vector \(\mathbf{f}(x)\)
|
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: \[\mathbf{f}(x) = \left\{\pm 1,\, \pm x,\, \pm x^2,\, \ldots,\, \pm x^n \right\} \tag{2.9}\]
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: \[\mathbf{f}(x) = \left\{\pm 1,\, \pm x,\, \pm x^2,\, \ldots,\, \pm x^n \right\} \tag{2.9}\]
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 :)
The materials used
- Wikipedia. A series of Taylor and Maclaurin.