How to fit a sinusoid to 2 points and their gradients

Question

Given the sinusoidal function

$$f(x) = a \cos(n x + b) + c,$$

if I know $f(x_1)$, $f(x_2)$, $f'(x_1)$ and $f'(x_2)$ is it possible to determine $a, b, c$ and $n$, with $x \in [0,\tfrac{2\pi}{n})$

Edit: put bounds on $x$ so that only one complete cycle is considered.

Edit 2: Current progress:

Let $p = p'/n$ and $x = x'-p'$ then

$$f(x) = a \cos(n x') + c$$

and

$$f'(x) = -an\sin(n x') + c$$

Let $x_2 = x_1 + w$, then we have

$$\frac{f'(x_2)}{f'(x_1)} = \frac{-na \sin(nx_1' + nw)}{-na \sin(nx_1')}$$ $$ = \cos(nw) + \frac{\sin(nw)}{\tan(nx_1')}$$

Rearranging, $$x_1' = \frac{\tan^{-1} \left( \frac{\sin(nw)}{\frac{f'(x_2)}{f'(x_1)} - \cos(nw)} \right)}{n}$$ which gives $p$ from before $(p = (x_1'-x_1)/n)$.

I'm most interested in finding $n$ so even if $a$ and $c$ can't be found it woudn't matter. Perhaps the above can be rearranged to give $n$?

Edit 2: Answered own question below.

Answer 1

Your question leads to the solvability of $\mathbf{F}(\mathbf{x},\mathbf{y})=0$ where $$\mathbf{x}=(a,n,b,c),$$ $$\mathbf{y}=(f_{1},f_{2},f'_{1},f'_{2}),$$ and $$\mathbf{F}(\mathbf{x},\mathbf{y})=\left[\begin{array}{l}a\cos(nx_{1}+b)+c-f_{1}\\a\cos(nx_{2}+b)+c-f_{2}\\-an\sin(nx_{1}+b)-f'_{1}\\-an\sin(nx_{2}+b)-f'_{2}\end{array}\right]. $$

So whenever $$\det J\mathbb{F}(a,n,b,c)=\left|\frac{\partial(F_{1},F_{2},F_{3},F_{4})}{\partial(a,n,b,c)}\right|\neq0,$$ it will follow from the implicit function theorem that the 4-tuple $(a,n,b,c)$ is solvable in terms of the given parameters $(f_{1},f_{2},f'_{1},f'_{2})$, and this will give you the desired sinusoid (note that $\mathbb{F}$ is continuously differentiable with respect to $\mathbb{x}$, so the Jacobian exists, only that its non-vanishing needs to be verified for some $\mathbb{y}$).

On the other hand, from a practical point of view, solving the non-linear system does not appear to be easy in general. A numerical method like Newton iteration could be used though if you actually needed to calculate the parameters, but it doesn't sound like this is what you're after.

Answer 2

There are trivially many solutions if $n$ is not known.

e.g. consider $f'(x_1) = v$, $f'(x_2) = -v$, $f(x_1) = f(x_2) = 0$ where $x_1 = -x_2$.

Then we have

$$ a = \frac{f'(x_1)}{-n\sin(n x_1)}$$ $$ b = 0 $$ $$ c = -a \cos(n x_1) $$ $$ n < \frac{\pi}{2 |x_1|} $$

which will satisfies the constraints.