Can we parametrize the Fermat curve by $g(t)^n+g'(t)^n=1$ for already named $g(t)$?

Basically the Fermat curve is $x^n+y^n=1$.

For $n > 2$, can we parametrize it by $x=g(t),y=g'(t)$ for already named function $g(t)$ of complex argument?

i.e., $g(t)^n+g'(t)^n=1$.

Some of what I know:

For $n=2$, solution is $g(t)=\sin(t)$.

For $n=4$, Wolfram alpha appears to give functional equation.

For $n \ge 3$, SAGE gives functional equation with a lot of constants:

#n = 3
x == 1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*(-t^3 + 1)^(1/3)/t - 1)) + _C + 1/3*log((-t^3 + 1)^(1/3)/t + 1) - 1/6*log(-(-t^3 + 1)^(1/3)/t + (-t^3 + 1)^(2/3)/t^2 + 1),
g(x) == (-t^3 + 1)^(1/3)
#n = 5
x == 1/5*sqrt(5)*(sqrt(5) + 1)*arctan((sqrt(5) + 4*(-t^5 + 1)^(1/5)/t - 1)/sqrt(2*sqrt(5) + 10))/sqrt(2*sqrt(5) + 10) + 1/5*sqrt(5)*(sqrt(5) - 1)*arctan(-(sqrt(5) - 4*(-t^5 + 1)^(1/5)/t + 1)/sqrt(-2*sqrt(5) + 10))/sqrt(-2*sqrt(5) + 10) + _C - 1/10*(sqrt(5) + 3)*log(-(-t^5 + 1)^(1/5)*(sqrt(5) + 1)/t + 2*(-t^5 + 1)^(2/5)/t^2 + 2)/(sqrt(5) + 1) - 1/10*(sqrt(5) - 3)*log((-t^5 + 1)^(1/5)*(sqrt(5) - 1)/t + 2*(-t^5 + 1)^(2/5)/t^2 + 2)/(sqrt(5) - 1) + 1/5*log((-t^5 + 1)^(1/5)/t + 1),
g(x) == (-t^5 + 1)^(1/5)

• What do you mean by "of complex argument"? Oh, and $g'$ really does mean derivative wrt $t$? – Per Alexandersson Oct 24 '16 at 16:27
• @PerAlexandersson Yes, it means derivative, isn't this standard notation? I mean that $t$ is complex and $g$ is function and $g'$ is the derivative of $g$. – joro Oct 24 '16 at 16:33
• Sure, it is standard notation - but I've seen people use the prime to indicate "dual" or complement as well, so just wanted to clarify... – Per Alexandersson Oct 24 '16 at 16:36
• Given that the Fermat curve is typically irrational (for $n>2$), can we parametrize it at all? – Alex Degtyarev Oct 24 '16 at 16:57
• @AlexDegtyarev Yes, we can parametrize it by (t, (1-t^n)^(1/n)) no matter of rationality. But this is not of the form (g(t),g't(t)) I am asking about. – joro Oct 24 '16 at 17:19

Let me give a perspective extending what Geoff did. He notes the function is the inverse function of the integral

$$\int_0^t \frac{1}{(1-u^n)^{1/n}} du$$

with the change of variables $v=u^n$ so $dv= n u^{n-1} du = n v^{(n-1)/n} dv$

$$= \frac{1}{n} \int_0^{t^{1/n}} \frac{1}{ (1-v)^{1/n} v^{(n-1)/n}} dv$$

we recognise the important special case $t=1$ of this integral as $1/n$ times the beta function $\beta((n-1)/n, 1/n)$.

In general the integral from $0$ to $t$ may depend on the choice of contour. The dependence is governed by a sheaf cohomology group, which here is one-dimensional. This should imply that the integral is well-defined up to the period $\beta((n-1)/n,1/n)$ times an element of $\mathbb Z[\mu_n]$.

In particular for $n \leq 4$, the integral is well-defined modulo a lattice in $\mathbb C$, so the inverse function (your $g(t)$) should be a standard elliptic function, periodic with respect to that lattice.

But for $n\geq 5$, the periods form a dense subset of $\mathbb C$, so the inverse function cannot be well-defined (the level sets of a well-defined holomorphic function are discrete) and in fact fails quite badly, taking infinitely many different values at each point. So this might explain why this function is not well-studied.

• Thanks for this insight Will. I was thinking in terms of functions of a real variable in my construction. What information does your explanation give about the radius of convergence of the putative Maclaurin series for $g$? – Geoff Robinson Oct 25 '16 at 14:12
• Thanks. So Geoff's answer is related to Weierstrass P function even though the models are different? And would it matter if for some other curve of positive genus the solution g(t) is from known functions? – joro Oct 25 '16 at 14:24
• @joro One interesting case of parameterisation of curves of positive genus by known functions is the parameterisation of modular curves by modular functions. However I don't think these usually satisfy a simple differential equation. – Will Sawin Oct 25 '16 at 14:43
• @GeoffRobinson That's a good question, though I'm not sure if this explanation is the best way to go about it. One wants to know the shortest path in $\mathbb C$ starting from $0$ such that integrating the differential equation $g^n+g'^n=1$ along that path reaches a point at $\infty$ of the Fermat curve. The endpoints of such paths form a coset of the group of periods, so for $n>4$ these are dense. But of course the length of the path is different from the distance from the endpoint to the origin. – Will Sawin Oct 25 '16 at 14:49
• @GeoffRobinson What i would try to understand is how the function $g(t)$ behaves on straight line paths out of the origin for every angle, and for which angles it reaches a singularity at the smallest radius. – Will Sawin Oct 25 '16 at 14:52

(This answer doesn't address the question for the whole curve, but really just the question of solving the differential equation near $0$).

I think one way to proceed is as follows, which can be made rigorous and is how things work when $n = 2:$ let $f(t) = \int_{0}^{t} \frac{1}{(1- u^{n})^{1/n}}du$ for $t \in (-1,1),$ where we take the unique real positive $n$-th root, and let $g$ be the inverse function to $f,$ which exists as $f$ is increasing where defined. Note that $f(0) = 0$ and that $f$ is an odd function if $n$ is even. Note also that (by the Fundamental Theorem of Calculus) $f$ is infinitely differentiable on $(-1,1).$

Then $f(g(t)) = t.$ Now $g^{\prime}(t) = \frac{1}{f^{\prime}(g(t))}$ and $f^{\prime}(g(t)) = \frac{1}{(1-g(t)^{n})^{\frac{1}{n}}}.$ Hence we do have $g(t)^{n} + g^{\prime}(t)^{n} = 1.$ Note also that $g(0) = 0$ and $g^{\prime}(0) = 1.$ From this, and the equation $g(t)^{n} + g^{\prime}(t)^{n} = 1,$ we can inductively determine the Maclaurin series for $g$.

Later edit partially motivated by Christian Remling's comment. The only difficulty seems to be to find ( near enough $0$) a good choice of $f$ in the complex case, and show that it is injective in neighbourhood of $0$: For the sake of completeness, I outline a proof that in the complex case, we can construct a solution $g(z)$ with $g(z)^{n} + g^{\prime}(z)^{n} =1,$ and $g(0) = 0,g^{\prime}(0) = 1,$ and that $g(z)$ has a Maclaurin series with strictly positive radius of convergence.

We define $f$ on the open unit disk $C(0;1)$ by $f(z) = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{1+nk}z^{1+nk} \left( \begin{array}{clcr} \frac{-1}{n}\\ k \end{array} \right).$ The power series does converge on $C(0;1),$ and we have $f^{\prime}(z) = (1-z^{n})^{\frac{-1}{n}} = \sum_{k=0}^{\infty} (-1)^{k}z^{nk} \left( \begin{array}{clcr} \frac{-1}{n}\\ k \end{array} \right)$ for $z \in C(0;1).$

Let $r$ be a real number with $0 < r < (1-\frac{1}{n+1})^{\frac{1}{n}}.$ Then $0 < \frac{r^{n}}{n(1-r^{n})} < 1.$ It is easy to check from the power series that for $z,w \in C(0;r),$ we have

$|z-w| ( 1 - \frac{r^{n}}{n(1-r^{n})}) \leq |f(z)-f(w)| \leq |z-w| ( 1 + \frac{r^{n}}{n(1-r^{n})}).$ In particular, $f$ is injective on $C(0;r),$ so the inverse function, $g$ say, is defined on the set $f(C(0;r)),$ which is an open set by the Open Mapping Theorem. The above inequalities imply that $g$ is uniformly continuous on $f(C(0;r)),$ and then it follows that $g^{\prime}(z) = \frac{1}{f^{\prime}(g(z))} = (1-g(z)^{n})^{\frac{1}{n}}.$ Hence we do have $g(z)^{n} + g^{\prime}(z)^{n} = 1$ for $z \in f(C(0;r)).$ Since $g$ is differentiable on an open set which includes $0,$ its Maclaurin series has a strictly positive radius of convergence.

We also note that for any complex $n$-th root of unity, we have $f(\omega z ) = \omega f(z)$ for any $z \in C(0;r)$ and for any (not necessarily primitive) $n$-th root of unity $\omega,$ so that we also have $g(\omega z) = \omega g(z)$ for any such $\omega$ and any $z \in f(C(0;r)).$ It follows easily from this that the Maclaurin series for $g(z)$ has the form $z + \sum_{k=1}^{\infty} a_{k} z^{1+nk}.$

• So to finish this remains to find the inverse function? – joro Oct 25 '16 at 5:36
• Well, the inverse function exists, but it may not be describable easily in terms of "already -known functions. – Geoff Robinson Oct 25 '16 at 12:59
• Likely you know this. SAGE solved the integral for n=3 and 4 with elementary functions. Wolfram Alpha said the inverse is not from known functions. WA's result heavy depends on what functions it knows and lack of bugs. – joro Oct 25 '16 at 13:45
• I saw in your question what you said about SAGE. I did not know that you had got that outcome from WA, but it would not surprise me that that is the case. – Geoff Robinson Oct 25 '16 at 13:56
• Isn't it obvious that $g$ is holomorphic near $t=0$, since $(1-z^n)^{1/n}$ is and the same basic theorems hold for complex $t$ (as for real $t$)? – Christian Remling Oct 27 '16 at 1:23

Suppose $u,v:\mathbb{C} \to \mathbb{C}$ are entire functions such that $u^{n}+v^{n}=1$ (disregarding relations like $u^{'}=v$ momentarily). Then $\frac{u}{v}$ is a meromorphic function which never attains a value among the $n$ solutions to $z^{n}=−1$. Picard's Little Theorem says that a meromorphic function omitting 3 values from its range is constant. Then $\frac{u}{v}$ is constant so $u$ and $v$ are themselves constant, or $n \leq 2$.

There is a general method to reparametrise ANY curve $(x(s),y(s))$ in the form $(g(t),g'(t))$ as required. We use the simple reparametrisation $$t=\int^s \dfrac {x'(u)}{y(u)} du.$$ In your case, we can employ the parametrisation $(\cos^{2/n}(s),\sin^{2/n}(s))$ as starting point.

Remarks:

1. In the general case, one has, of course, to avoid points where the curve crosses the $x$-axis. In the cases in point here one can restrict attention to the part of the curve which lies in the first quadrant and then use the usual symmetry argument. One can then use mathematica to attack the resulting integrals (I have no access at the moment).This, of course, leaves open the question of an explicit parametrisation.It will depend on the value of $n$ whether the integral and the functional inversion required to get $s$ as a function of $t$ can be carried out explicitly.

2. I thought that it would be of interest to the OP to see his problem embedded into a more general situation. For a more careful treatment of how to obtain such a parametrisation for a general curve, see the arXiv paper 1102.1579. This also has many examples which show the relevance and usefulness of such parametrisations.