I was playing around a bit this morning until I realized that one could write the arbitrary derivative of $\cos x$ as

$$(\cos x)^{(n)} = \cos\big(\frac{\pi}{2}(n-1) \big)\sin x + \cos\big(\frac{\pi}{2}n\big)\cos x$$

Similarly, writing $$\sin x=\cos \left(x-\frac{\pi}{2}\right)$$

we see that $$\frac{d^n}{dx^n}\sin x = \frac{d^n}{dx^n} \cos \left(x-\frac{\pi}{2}\right)$$

which evaluates to (I think I'm using the chain rule properly in this case):

$$\cos\big(\frac{\pi}{2}(n-1) \big)\sin \left(x-\frac{\pi}{2}\right) + \cos\big(\frac{\pi}{2}n\big)\cos\left(x-\frac{\pi}{2}\right)$$


$$(\sin x)^{(n)} = \cos\big(\frac{\pi}{2}(n-1) \big)\cos x + \cos\big(\frac{\pi}{2}n\big)\sin x$$

where we have that $n \in \mathbb{R}$

These two results can be expressed as a complex exponential too.

Because these functions happens to be continuous, in some ways the differential operator is now continuous for cosine and sine.

I'm wondering if anyone can interpret this for me a I haven't studying real analysis and only have a nonrigorous calculus background.


After doing some reading, I realized that because every function can be written as a Fourier transform, one ought to be able to simply use the above definitions to take the nth derivative of the fourier transform of an arbitrary function to an arbitrary degree of precision (by choosing how many terms to include).

Is the hypothesis true? If so, can you summarize it in a way I would be able to understand and use. Assume I know the definition of the fourier transform. All I am looking for now is a general equation for the continuous differential operator of a function.

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    $\begingroup$ +1 for curiousity. See en.wikipedia.org/wiki/Fractional_calculus $\endgroup$ Mar 22, 2017 at 12:31
  • $\begingroup$ Thanks so much. I did think about this many years ago when I hadn't even learned calculus yet and only had an intuitive notion of what the derivative is. Now that I know calculus, I tried to find the continuous derivative of cosine. However, whenever I periodically revisit that page I understand nothing. $\endgroup$ Mar 22, 2017 at 12:35
  • $\begingroup$ I just realized that this expression looks very similar to $f(x) = y=\mbox{Re}\left( \cos \left( \frac{x\pi }{2} \right)+i\sin \left( \frac{x\pi }{2} \right) \right)$. Is there a connection? $\endgroup$ Mar 22, 2017 at 13:04
  • $\begingroup$ I'm not an expert on the topic so can't really help with your particular question. I just wanted to point out that the idea of a fractional derivative is well known. $\endgroup$ Mar 22, 2017 at 13:08
  • $\begingroup$ Got it. Thanks so much. I'll wait for someone to answer $\endgroup$ Mar 22, 2017 at 13:09

1 Answer 1


You have a set of functions $\{ \sin x, \cos x \}$ that is closed under differentiation because $\sin ' = \cos$ and $\cos' = -\sin$. So you are restricting differentiation to the two-dimensional subspace $M$ generated by these functions, which consists of all $f=A\cos x + B \sin x$ with constants $A$ and $B$. In this context $\frac{d}{dx}$ is a linear operator on a two-dimensional vector space. No matter what norm you put on this space, the topology will be the same because of it being finite dimensional. So $\frac{d}{dx}$ will be continuous no matter what norm you choose for this subspace of functions. You can do the same with $\{ \cos(nx),\sin(nx) \}$ for $n=1$ or $n=2$, or $n=3$, and you can combine any number of these as well. You'll end up with a space of functions consisting of linear combinations of such functions, and differentiation will map that space back to itself.

The matrix representation of $\frac{d}{dx}$ acting on a basis $\{\cos(nx),\sin(nx)\}$ has the form $$ \left[\begin{array}{cc}0 & -n \\ n & 0\end{array}\right]=n\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right] $$ which is $n$ times a counter-clockwise rotation by $90^o$. The $k$th power of a such a matrix can be written as $n^k$ times a rotation by $k \times 90^o$, which leads to a natural extension to a real power $r$ which is $n^r$ times a rotation about $r \times 90^o$: $$ n^{r}R(r) = n^{r}\left[\begin{array}{cc} \sin(\frac{\pi}{2}r) & -\cos(\frac{\pi}{2}r) \\ \cos(\frac{\pi}{2}r) & \sin(\frac{\pi}{2}r) \end{array}\right] $$

Multiplication by $i$ also serves to rotate counterclockwise in the complex plane by $90^o$, which has a natural extension to multiplication by $e^{ir\theta}$, which is why the exponential power of the trigonometric functions is a bit easier to deal with.

In terms of the exponential Fourier series $$ \frac{d^r}{d\theta^r}e^{in\theta}=(in)^re^{in\theta} = (ne^{i\pi/2})^re^{in\theta} = n^r e^{ir\pi/2}e^{in\theta} $$ So, $$ \frac{d^r}{d\theta^r}\sum_{n=-\infty}^{\infty}c_n e^{in\theta} = \sum_{n=-\infty}^{\infty}n^r e^{i\pi r/2} e^{in\theta}. $$ This is a reasonable definition of fractional derivatives for $r > 0$ that agrees with the usual definition for $r=1,2,3,\cdots$.

The same sort of thing works for the Fourier transform, too: \begin{align} \frac{d^r}{dx^r}f & = \frac{d^r}{dx^r} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{ixs}\hat{f}(s)ds \\ & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(is)^r e^{isx}\hat{f}(s)ds. \end{align} These ideas fall under the general heading of Functional Calculus for operators.

  • $\begingroup$ Judging from the comments to the original question, this is not the kind of "continuity" the OP has in mind. I think that the OP is about interpreting the parameter $n$ in the formulas for $\frac{d^n}{dx^n}$ as a continuous (=real) parameter. $\endgroup$ Mar 26, 2017 at 17:13
  • $\begingroup$ @GiuseppeNegro : Now that you say that, I'm not sure. $\endgroup$ Mar 26, 2017 at 17:44
  • $\begingroup$ I mean that the differentiation operator itself is now defined for all real numbers for any uniformly convergent function $\endgroup$ Mar 26, 2017 at 18:29
  • $\begingroup$ @theideasmith : Okay. I've added a little more for you from that point of view. $\endgroup$ Mar 26, 2017 at 19:08
  • $\begingroup$ Thanks so much! Thinking about the derivative in terms of a vector space is quite enlightening. Can you also comment on extending this result to all functions by use of the fourier transform $\endgroup$ Mar 26, 2017 at 19:09

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