Representing the cyclic differentiation pattern of $\frac{d^n}{dx^n}(\sin(x))$ using linear algebra.

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 derrivative). 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 derrivative 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 sidenote I am using the equivalence notation loosely so if there is a better way to write it let me know.

Answer 1

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

• How did I come to 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 same 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 $90^0$ that is $\hat i \text { lands on } \hat j$ thus 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 $n$th time we have $$ \pmatrix{\frac{d^nS}{dx^n} \\ \frac {d^nC}{dx^n}} =\pmatrix{0 & -1 \\ 1 & \ 0 }^n\pmatrix{S \\ C }$$

Answer 2

Write $$ \pmatrix{\cos' \\ \sin'} = \pmatrix{0 & -1 \\ 1 & \hphantom{-}0} \pmatrix{\cos \\ \sin } $$ Let the matrix above be $J$. Then $J^2=-I$. In fact, $J$ corresponds to $i$ in the matrix representation of complex numbers. And you can take $J^t$ for any real number $t$ by considering rotations by $t\pi/2$.