Forget everything you know about $\sin$ and $\cos$. Forget everything about $\sinh$ and $\cosh$. Forget about elliptic integrals and elliptic functions. The goal here is to discover a power series through discovery.

Suppose we know about differentiability, power series, and computing lengths of continuously differentiable paths. Suppose we wanted to discover two functions $x$ and $y$ (written as functions of $t$) which essentially "wrap the real line around the unit circle in an isometric fashion" which sends $0$ to $(1,0)$ and sends all sufficiently small positive numbers into the first quadrant. Suppose we are lucky enough that $x$ and $y$ are continuously differentiable. Then $x$ and $y$ would have the following properties:

$x^2+y^2=1\quad$ (unit circle condition)

$\dot{x}^2+\dot{y}^2=1\quad$ (isometric condition)

$x(0)=1\qquad\text{and}\qquad y(0)=0\quad$ (normalization condition)

$y$ is locally strictly-increasing around $0\quad$ (orientation condition)

$x$ has a local maximum at $0\quad$ (included to make some work easier, possibly not needed)

Through a bunch of gymnastics, one can show that $x$ and $y$ are both in fact $C^\infty$ and that they are both solutions to the differential equation $$f''+f=0$$ Using the initial conditions and the differential equation, we can compute $x$ and $y$'s Taylor series as $$x(t)=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}t^{2k}\qquad\text{and}\qquad y(t)=\sum_{k=0}^\infty\frac{(-1)^{2k+1}}{(2k+1)!}t^{2k+1}$$ We can then show that these Taylor series converge for all real $t$, and that these series actually converge to $x$ and $y$ respectively.

I like this approach because it starts off with a very concrete construction problem, and we discover the two functions based upon few assumptions. I want to attempt the same thing with "wrapping the real line onto the right-branch of the unit hyperbola". This time the conditions change:



$x(0)=1\quad\text{and}\quad y(0)=0$

$y$ is strictly increasing

$x$ has a local minimum at $0$

Through a few hoops, we arrive at two nonlinear differential equations $$(\dot{x})^2=\frac{x^2-1}{2x^2-1}\qquad\text{and}\qquad(\dot{y})^2=\frac{y^2+1}{2y^2+1}$$ I am having difficulty manipulating all the information here to find a differential equation (namely, a linear differential equation) that $x$ and $y$ solve from which I can compute a Taylor series fairly easily. Somewhat unsurprisingly, a symbolic algebra system will return an answer involving a hyperbolic function and elliptic functions/integrals. I don't care about that. An ideal answer here will work from the first principles here to compute a Taylor series and fill in the work (it could just be an outline) as to how they were arrived at. Arguably, one could use the differential equations I have already found and use algebra of power series to compute the coefficients. I have taken this route and have found the computations tedious and time-consuming. Another answer I would appreciate is demonstrating a sleek way to make that kind of computation less tedious.

Any help is appreciated.

Update Since $x$ and $y$ turn out to involve elliptic functions, I doubt these functions satisfy a linear differential equation. However, I would still appreciate an approach demonstrating a sleek way to compute the Taylor coefficients. Since we are assuming isometricity, we can also assume that $x$ is an even function and that $y$ is an odd function of $t$. This might make an approach slightly easier.

  • $\begingroup$ suggest $\dot{y}^2-\dot{x}^2=1$ for the hyperbola. $\endgroup$
    – Will Jagy
    Aug 13, 2016 at 21:00
  • $\begingroup$ @WillJagy Are you suggesting I show that? I agree that $\dot{y}^2-\dot{x}^2=(x^2+y^2)^{-1}$, but we don't know that $x^2+y^2=1$ for the hyperbola. $\endgroup$
    – user123641
    Aug 14, 2016 at 1:47
  • $\begingroup$ @Bryan: I don't understand: what do you mean by "wrap the real line around the unit circle in an isometric fashion"? The circle and the straight line cannot be locally-isometric, because they don't have the same curvature. $\endgroup$
    – Alex M.
    Aug 18, 2016 at 16:38

1 Answer 1


You probably know that one possible parametrization for the right component of the hyperbola $x^2 - y^2 = 1$ is $t \mapsto (\cosh t, \sinh t)$. The tangent vector at this trajectory is $(\dot x, \dot y) = (\sinh t, \cosh t)$, and its components do not satisfy the relation $\dot x^2 + \dot y^2 = 1$ that you wish to impose, but rather $\dot x ^2 - \dot y ^2 = -1$. This has to do with the fact that, when discussing the hyperbola, the natural ambient space in which to view it is not the $2$-dimensional Euclidean space, but rather $\Bbb R^2$ endowed with the pseudo-Euclidean product $\langle (x,y), (u,v) \rangle = -xu + yv$ (and the corresponding quadratic form $q \big( (x,y) \big) = -x^2 + y^2$). (Notice that I have chosen the opposite convention to the one used on Wikipedia, but this does not alter the underlying ideas.)

This means that the right component of the hyperbola is $q ^{-1} (-1)$, and that, if $\gamma (t) = \big( x(t), y(t) \big)$, your equations can be rewritten as

$$\begin{align} q \big( \gamma (t) \big) &=& -x ^2 + y ^2 &=& -1 \\ q \big( \dot \gamma (t) \big) &=& - \dot x ^2 + \dot y ^2 &=& 1 \\ q \big( \gamma (0) \big) &=& \big( x(0), y(0) \big) &=& (1, 0) . \end{align}$$

In differential geometrical parlance, the middle equation says that $\gamma$ is "parametrized with unit speed", or "canonically". Notice that I shall not need the local minimum condition on $x$, but I shall use the monotonicity condition on $y$ (I could have done the opposite, the idea being that a single of these conditions is needed - the underlying idea being that any of them helps us choose one of the two possible orientations on the curve).

Deriving the first two equations with respect to $t$ gives

$$\langle \dot \gamma, \gamma \rangle = - \dot x x + \dot y y = 0 \\ \langle \dot \gamma, \ddot \gamma \rangle = - \ddot x \dot x + \ddot y \dot y = 0 .$$

If $0 \ne v \in \Bbb R^2$, denote by $v ^\perp$ its orthogonal complement with respect to the pseudo-Euclidean product introduced above. What can the dimension of $v ^\perp$ be? Well, since it must be a linear subspace of $\Bbb R^2$, its dimension may be only $0$, $1$ or $2$.

  • If $\dim v ^\perp = 0$, then the only vector perpendicular to $v$ is $0$. This is not possible, though, because if $v = (a,b)$ then $(b, a)$ is orthogonal to it and non-zero: $\langle (a,b), (b, a) \rangle = -ab + ba = 0$.

  • If $\dim v ^\perp = 2$, then the whole $\Bbb R^2$ would be perpendicular to $v \ne 0$, but this would mean that the pseudo-Euclidean product considered is degenerate, which it clearly is not (its matrix is $\begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix}$, having determinant $-1 \ne 0$)

  • $v ^\perp$ being non-empty (it includes at least $(b,a)$ as discussed above), it follows that it must have dimension $1$.

The two equations written above say that $\gamma, \ddot \gamma \in \dot \gamma ^\perp$, and since $\dim \dot \gamma ^\perp = 1$, it follows that there exist $k \in \Bbb R$ such that $\ddot \gamma = k \gamma$, i.e. we have the differential equations

$$\ddot x = k x \\ \ddot y = k y .$$

If $k = 0$, you get no solution satisfying all the imposed conditions, so $k \ne 0$.

Assuming that

$$x = \sum _{n \ge 0} a_n t^n, \quad y = \sum _{n \ge 0} b_n t^n$$

(an assumption that is justified, a posteriori, by the fact that it allows us to find the unique solution), inputting this into the differential equations leads to

$$\sum _{n \ge 0} (n+2) (n+1) a_{n+2} t^n = \sum _{n \ge 0} k a_n t^n, \quad a_0 = 1 \\ \sum _{n \ge 0} (n+2) (n+1) b_{n+2} t^n = \sum _{n \ge 0} k b_n t^n, \quad b_0 = 0 ,$$

which in turn (use a bit of induction, but it will be trivial) leads to

$$\begin{align} & a_0 = 1, \ a_{2n} = \frac k {(2n)!} \ n \ge 1, \quad & a_{2n+1} = \frac k {(2n+1)!} a_1 \\ & b_{2n} = 0, \quad & b_{2n+1} = \frac k {(2n+1)!} b_1 , \end{align}$$

which is the same as

$$\begin{align} & x = 1 + k \sum _{n \ge 1} \frac {t^{2n}} {(2n)!} + k a_1 \sum _{n \ge 0} \frac {t^{2n+1}} {(2n+1)!} \\ & y = k b_1 \sum _{n \ge 0} \frac {t^{2n+1}} {(2n+1)!} . \end{align}$$

Notice that the series of $y$ contains only odd powers, therefore $y$ must be an odd function, so

$$x(t) ^2 = y(t) ^2 + 1 = \big( - y(-t) \big) ^2 + 1 = y(-t) ^2 + 1 = x(-t) ^2 .$$

Let's impose this condition, then:

$$\left( 1 + k \sum _{n \ge 1} \frac {t^{2n}} {(2n)!} + k a_1 \sum _{n \ge 0} \frac {t^{2n+1}} {(2n+1)!} \right) ^2 = \left( 1 + k \sum _{n \ge 1} \frac {(-t)^{2n}} {(2n)!} + k a_1 \sum _{n \ge 0} \frac {(-t)^{2n+1}} {(2n+1)!} \right) ^2 = \\ \left( 1 + k \sum _{n \ge 1} \frac {t^{2n}} {(2n)!} - k a_1 \sum _{n \ge 0} \frac {t^{2n+1}} {(2n+1)!} \right) ^2 ,$$

so squaring and reducing terms gives

$$\left( 1 + k \sum _{n \ge 1} \frac {t^{2n}} {(2n)!} \right) \ k a_1 \sum _{n \ge 0} \frac {t^{2n+1}} {(2n+1)!} = 0 \ \forall t ,$$

which implies $k a_1 = 0$ which, since $k \ne 0$, implies $a_1 = 0$, so that

$$x = 1 + k \sum _{n \ge 1} \frac {t^{2n}} {(2n)!} .$$

By visual inspection of the obtained series, we recognize that

$$x = 1 + k (\cosh t - 1), \quad y = k b_1 \sinh t .$$

Next, using $- \dot x ^2 + \dot y ^2 = 1$, we obtain

$$1 = - k^2 \sinh ^2 t + k^2 b_1 ^2 \cosh^2 t = -k^2 (\cosh^2 t -1) + k^2 b_1 ^2 \cosh^2 t = k^2 (b_1 ^2 - 1) \cosh^2 t + k^2 \ \forall t ,$$

which implies $b_1 ^2 = 1$ and $k^2 = 1$.

Since $y$ is required to be strictly increasing (this is the only place where I use this), we must also have $k b_1 > 0$, therefore there are only two possibilities: $b_1 = k = 1$ and $b_1 = k = -1$.

The first possibility gives $x = \cosh t, \ y = \sinh t$. The second one gives $x = 2 - \cosh t, \ y = \sinh t$, which doesn't satisfy $-x^2 + y^2 = -1$ for all $t$.

To conclude, then, the only solution to your problem is

$$x = \sum _{n \ge 0} \frac {t^{2n}} {(2n)!} = \cosh t , \\ y = \sum _{n \ge 0} \frac {t^{2n+1}} {(2n+1)!} = \sinh t .$$

  • $\begingroup$ Strictly speaking, this isn't an answer to my question, but I appreciate that you show how the canonical parameterization of the hyperbola is 'natural' with respect to a different inner product. $\endgroup$
    – user123641
    Aug 24, 2016 at 13:10

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