Question

It is known that the derivatives of $\sin(x)$ and $\cos(x)$ follow this pattern:

$$\frac{d^1}{dx^1}(\sin(x))=\cos(x)$$ $$\frac{d^2}{dx^2}(\sin(x))=-\sin(x)$$ $$\frac{d^3}{dx^3}(\sin(x))=-\cos(x)$$ $$\frac{d^4}{dx^4}(\sin(x))=\sin(x)$$

Motivated by the matrix representation of complex numbers, I thought one could interpret this pattern in the context of a coordinate plane and 90-degree rotations ($\sin$ rotates around to cosine, rotates around again to negative $\sin$, and so on for each derivative). One can assign the following basis vectors for the space:

$$\mathbf{e}_{1}=\sin(x);\quad\mathbf{e}_{2}=\cos(x)$$

Thus, a linear combination of a sine and cosine function relates to a vector in the following way:

$$C_1\sin(x)+C_2\cos(x)\equiv\begin{pmatrix} C_1 \\ C_2 \end{pmatrix}$$

The \( n \)'th derivative can be interpreted as the following matrix-vector product:

$$\frac{d^n}{dx^n}(C_1\sin(x)+C_2\cos(x))\equiv\begin{pmatrix} \cos(\frac{n\pi}{2}) & -\sin(\frac{n\pi}{2}) \\ \sin(\frac{n\pi}{2}) & \cos(\frac{n\pi}{2})\end{pmatrix}\begin{pmatrix} C_1 \\ C_2 \end{pmatrix}$$

So taking a derivative is equivalent to rotating the vector representing the function by 90 degrees. I believe that the inner product would then be defined as:

$$\langle C_1\sin(x)+C_2\cos(x),C_3\sin(x)+C_4\cos(x)\rangle=C_1C_3+C_2C_4$$

Though this doesn't seem to have an immediately useful interpretation. If you let \( n \in \mathbb{R} \), then (using the sum of angles formula) a possible value for the fractional derivative of \( C_1\sin(x)+C_2\cos(x) \) is given by:

$$\frac{d^{n}}{dx^{n}}(C_1\sin(x)+C_2\cos(x))=C_1\sin(x+\frac{n\pi}{2}) + C_2\cos(x+\frac{n\pi}{2})$$

For \( f(x)=C_1\sin(x) \) and \( n=\frac{1}{2} \), one obtains:

$$\frac{d^{0.5}}{dx^{0.5}}(C_1\sin(x))=C_1\sin(x+\frac{\pi}{4})$$

Which coincides with the answer given here. Has such a representation of the derivative of $\sin(x)$ been investigated before? What useful applications can be obtained from this interpretation? As a side note, I am using the equivalence notation loosely, so if there is a better way to write it, let me know.

Answer

$$ \pmatrix{\frac{d^nS}{dx^n} \\ \frac {d^nC}{dx^n}} =\pmatrix{0 & -1 \\ 1 & \ 0 }^n\pmatrix{S \\ C } $$

How did I come to the above formula?

We can use the change of basis of vectors (thanks to $3$Blue$1$Brown YT channel).

We know $\frac {d^nS}{dx^n} = S(x + \frac {n\pi}{2})$ and similarly for $\frac {d^nC}{dx^n} = C(x + \frac {n\pi}{2})$.

Here, we note the fact that every time we differentiate, the result we get is the rotating axis with $90^\circ$, that is, $\hat i \text{ lands on } \hat j$, and the same keeps happening every time we are differentiating.

$$ \pmatrix {\cos\theta & \sin\theta\\-\sin\theta&\cos\theta}_{\theta = \frac\pi2} = \pmatrix{0 & -1\\1 & 0} $$

Thus, for the \( n \)th time, we have:

$$ \pmatrix{\frac{d^nS}{dx^n} \\ \frac {d^nC}{dx^n}} =\pmatrix{0 & -1 \\ 1 & \ 0 }^n\pmatrix{S \\ C } $$