For \(n\geq 2\) we can define a curve \(z:[0,2\pi]\to\mathbb{C}\) by \(z(t):=ne^{it}-e^{int}\). It is the path that a point on a circle of radius \(1\) follows, when that circle is rolled around a circle of radius \(n-1\). In the examples below, we let \(n=4\).
Using Euler’s formula, we can expand \(e^{int}=(\cos(t)+i\sin(t))^n\). Let \(z=x+iy\). Then \(x(t)\) and \(y(t)\) are polynomials in \(\cos(t)\) and \(\sin(t)\).
\[\begin{align*} x(t) &= -\cos^4(t) + 6\cos^2(t)\sin^2(t) - \sin^4(t) + 4\cos(t) \\ y(t) &= -4\cos^3(t)\sin(t) + 4\cos(t)\sin^3(t) + 4\sin(t) \end{align*}\]
Using the identity \(\sin^2(t)=1-\cos^2(t)\) we can rid of the \(\sin\)s in \(x\). That won’t work for \(y\) because \(y\) is an odd function, but we can do it for \(y^2\) instead. Using the abbreviation c for \(\cos(t)\) we get two functions, f and g, that only depend on c:
f(c) = -8*c^4 + 8*c^2 + 4*c - 1 g(c) = -64*c^8 + 128*c^6 + 64*c^5 - 80*c^4 - 96*c^3 + 32*c + 16
The resultant of two polynomials is defined as the determinant of a certain matrix involving their coefficients. It is zero if and only if the polynomials have a common root. If we now let \(x\) and \(y\) denote any given real numbers, the statement \[\text{Res}_c(x-f(c),y^2-g(c))=0\] is true if and only if there exists a \(c\) such that \(x=f(c)\) and \(y^2=g(c)\). Computing the resultant therefore gives an implicit equation for the epicycloid: \[\begin{array}{c}256x^8 + 16384x^6y^2 + 393216x^4y^4 + 4194304x^2y^6 + 16777216y^8 \\ - 5120x^6 - 245760x^4y^2 - 3932160x^2y^4 - 20971520y^6 - 2560x^4 \\ - 81920x^2y^2 - 655360y^4 + 131072x^3 - 6291456xy^2 - 46080x^2 \\ - 737280y^2 - 864000 = 0\end{array}\]
Using SageMath:
n = 4; var('c','s','x','y'); assume(c,"real"); assume(s,"real"); z = n*(c + I*s) - (c + I*s)^n; f = expand(real(z)); g = expand(imag(z)^2); dict = {s^i : (1 - c^2)^(i/2) for i in range(2,g.degree(s)+1,2)}; f = expand(f.subs(dict)); g = expand(g.subs(dict)); expand((x - f).resultant(y^2 - g,c))
The multibrot set is defined as the set of all complex numbers \(c\) such that \(P_c^k(0)\) remains bounded, where \(P_c(z)=z^n+c\). It contains an epicycloid about \(0\).
The cusp on the positive real axis is at the greatest \(c\) such that \(P_c^k(0)\) converges. By drawing a cobweb diagram, we see that the line \(y=x\) must be tangent to \(P_c\).
From \(P_c'(x)=1\) we get \(nx^{n-1}=1\), and from \(P_c(x)=x\) we get \(x^n+c=x\). So we have
\[x=\left(\frac1n\right)^\frac1{n-1}\text{ and then }c=x-x^n=\left(\frac1n\right)^\frac1{n-1}-\left(\frac1n\right)^\frac n{n-1}\]
Some values:
\[\begin{align*}\frac12-\left(\frac12\right)^2 &= 0.25 \\ \sqrt[3]{\frac14}-\sqrt[3]{\left(\frac14\right)^4} &= 0.47247\dots \\ \sqrt[9999]{\frac1{10000}}-\sqrt[9999]{\left(\frac1{10000}\right)^{10000}} &= 0.998979\dots\end{align*}\]