If one rotates a function such as the sine function about the origin, is there a general method to find the taylor series for the rotated function? Assuming of course that the rotated function is still a single valued function.

Rotating $y=f(x)$ counterclockwise by $\theta$ would give a function $g$ satisfying $g(x\cos\theta f(x)\sin\theta)=x\sin\theta+f(x)\cos\theta$, right? – Jonas Meyer Sep 30 '13 at 02:20

"Rotated function" doesn't necessarily have to be a function – Shuchang Sep 30 '13 at 02:33

@ShuchangZhang: PMay acknowledged that I think by adding the assumption that the rotated function is still a function. – Jonas Meyer Sep 30 '13 at 02:48

@JonasMeyer OIC – Shuchang Sep 30 '13 at 03:00
2 Answers
Setup before Setup:
[You can skip this if you want to just see an answer but this will help you develop the intuition to build these answers yourself]
Let's recall how we build taylor series but in a slightly more convoluted way (and this will be important because it will generalize very nicely to rotation)
The way it goes is you start with a function
$$ y = f(x)$$
Now the way we get the taylor series of this is to differentiate this thing a bunch of times and then fix a value $x=a$ and use that to assemble your taylor series
$$ y = f(x) \\ \frac{dy}{dx} = f'(x) \\ \frac{d^2 y}{dx^2} = f''(x) \\ \vdots $$
Now with our substitution (i know this is trivial but bear with me)
$$ y_a = f(a) \\ \frac{dy_a}{dx} = f'(a) \\ \frac{d^2 y_a}{dx^2} = f''(a) \\ \vdots $$
Now our taylor series formula is that:
$$ f(x) = y_a + \left(\frac{dy_a}{dx} \right) (xa) + \frac{1}{2!} \left(\frac{dy_a}{dx} \right) (xa)^2 + ... $$
Again we can make our substitutions $y_a = f(a), \frac{dy_a}{dx} = f'(a), ... $ etc to write this in the more familiar form
$$ f(x) = f(a) + \left(f'(a) \right) (xa) + \frac{1}{2!} \left(f''(a) \right) (xa)^2 + ... $$
Setup:
We can motivate how we understand the rotation problem by first considering the well known case of computing a taylor series for the inverse of a function.
So now we want to compute our taylor series for $f^{1}$. Suppose that $y = f^{1}(x)$ then we know that $$ f(y) = x $$
Now again we want to differentiate these a bunch of times and isolate $y$, we start by differentiating.
$$ f(y) = x \\ f'(y) \frac{dy}{dx} = 1 \\ f''(y) \left( \frac{d y}{dx} \right)^2 + f'(y) \frac{d^2y}{dx^2} = 0 \\ \vdots $$
Now Here is where it gets a bit messy, we actually have to solve for our terms. So we can start with the first two terms
$$ y = f^{1}(x) \\ \frac{dy}{dx} = \frac{1}{f'(y)} \\ f''(y) \left( \frac{d y}{dx} \right)^2 + f'(y) \frac{d^2y}{dx^2} = 0 \\ \vdots $$
Now this formula is weird because that second term depends on $y$ and that third term is also a little wonky cuz we have a $\left( \frac{dy}{dx} \right)^2 $ already floating in there so even after solving for $\frac{d^2 y}{dx^2}$ we then have to make substitutions...
$$ y = f^{1}(x) \\ \frac{dy}{dx} = \frac{1}{f'(f^{1}(x))} \\ \frac{d^2y}{dx^2} =  \frac{f''(y)}{f'(y)} \left( \frac{d y}{dx} \right)^2 =  \frac{f''(y)}{f'(y)^3} =  \frac{f''(f^{1}(x))}{f'(f^{1}(x))^3} \\ \vdots $$
Great so now that we have our derivatives we can make our taylor series, we just fix a choice of $x=a$ and roll from there
$$ f(x) = y_a + \left( \frac{dy_a}{dx} \right)(xa) + \frac{1}{2!} \left( \frac{d^2 y_a}{dx^2} \right)(xa)^2 + ... $$
I'm not going to do the substitution this time because it will be very messy and in fact its so ugly looking they even made a theorem to describe it in a more succinct way that doesn't break on edge cases
The general strategy:
So what actually happened here? We started with the expression $y = f(x)$ we considered the inverse of this function by making coordinate transformation $\left(x, y \right) \rightarrow (y, x)$ so that our formula $y = f(x)$ became instead $x = f(y)$. We then differentiated this $x = f(y)$ formula a bunch of times, isolated all the $\frac{d^ky}{dx^k}$ terms and then built our taylor series.
This procedure doesn't ONLY work on inverses. It works any time we can describe our coordinate transformation explicitly.
So we can now apply this to rotation. Rotation of a point $(x,y)$ about the origin by an angle $\theta$ counterclockwise can be given by the transformation: $$(x,y) \rightarrow (x \cos \theta  y \sin \theta, x \sin \theta + y \cos \theta)$$
To see where this comes from you can interpret $(x,y)$ as the column vector $\begin{bmatrix} x \\ y \end{bmatrix}$ and multiply it by the rotation matrix.
So now that we have our transformation we start with our original function
$$ y = f(x) $$
And we apply the transformation
$$ x \sin \theta + y \cos \theta = f\left( x \cos \theta  y \sin \theta \right) $$
And now you need to differentiate this a bunch of times, solve for $y, \frac{dy}{dx}, \frac{d^2y}{dx^2}, ... $ (as many terms as you like). And then assemble your taylor series, [unfortunately solving for $y$ itself seems difficult but we can do the first derivative]
$$ \sin \theta + \frac{dy}{dx} \cos \theta = f' \left( x \cos \theta  y \sin \theta \right) \left( \cos \theta  \frac{dy}{dx} \sin \theta \right) $$
$$ \frac{dy}{dx} = \frac{f' \left( x \cos \theta  y \sin \theta \right) \cos \theta  \sin \theta}{\cos \theta  \sin \theta f' \left( x \cos \theta  y \sin \theta \right)} $$
If you actually need to build this taylor series out I would recommend using some tricks such as not explicitly picking a point $x=a$ but rather fixing a particular value of $x \cos \theta  y \sin \theta $ so that these formulas become much simpler.
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I have limited time to post the answer, but I think this is it. The individual coefficients get multiplied such that $b_n = a_n\exp(2n\pi i k)$, where k is the rotation angle, some rational or irrational rotational coefficient scaled to be between zero and one, such that $\theta=\exp(2\pi i k)$
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Ok, take a very simple example. $f_1=x$. $f_2=ix$. I think $f_1$ is rotated by $\frac{\pi i}{2}$ with respect to $f_2$, if one looks at the unit circle for both function in the complex plane. Perhaps that's not the type of rotation the op was interested in, and perhaps someone else can define another type of rotation, and answer the op's question. – Sheldon L Nov 20 '13 at 16:35