Rose (mathematics)

A rose is the set of points in polar coordinates specified by the polar equation

${\displaystyle r=a\cos(k\theta )}$[2]

or in Cartesian coordinates using the parametric equations

${\displaystyle x=r\cos(\theta )=a\cos(k\theta )\cos(\theta )}$

${\displaystyle y=r\sin(\theta )=a\cos(k\theta )\sin(\theta )}$.

Roses can also be specified using the sine function.[3] Since

${\displaystyle \sin(k\theta )=\cos \left(k\theta -{\frac {\pi }{2}}\right)=\cos \left(k\left(\theta -{\frac {\pi }{2k}}\right)\right)}$.

Thus, the rose specified by ${\displaystyle \,r=a\sin(k\theta )}$ is identical to that specified by ${\displaystyle \,r=a\cos(k\theta )}$ rotated counter-clockwise by ${\displaystyle \pi /2k}$ radians, which is one-quarter the period of either sinusoid.

Since they are specified using the cosine or sine function, roses are usually expressed as polar coordinate (rather than Cartesian coordinate) graphs of sinusoids that have angular frequency of ${\displaystyle k}$ and an amplitude of ${\displaystyle a}$ that determine the radial coordinate ${\displaystyle (r)}$ given the polar angle ${\displaystyle (\theta )}$ (though when ${\displaystyle k}$ is a rational number, a rose curve can be expressed in Cartesian coordinates since those can be specified as algebraic curves[4]).

Roses are directly related to the properties of the sinusoids that specify them.

• A rose specified as ${\displaystyle r=a\cos(k\theta )}$ is symmetric about the polar axis because ${\displaystyle a\cos(k\theta )=a\cos(-k\theta )}$.

• A rose specified as ${\displaystyle r=a\sin(k\theta )}$ is symmetric about the vertical line ${\displaystyle \theta =\pi /2}$ because ${\displaystyle a\sin(k\theta )=a\sin(\pi -k\theta )}$.

• Graphs of roses are composed of petals. A petal is the shape formed by the graph of a half-cycle of the sinusoid that specifies the rose. (A cycle is a portion of a sinusoid that is one period ${\displaystyle T=2\pi /k}$ long and consists of a positive half-cycle, the continuous set of points where ${\displaystyle r\geq 0}$ and is ${\displaystyle T/2=\pi /k}$ long, and a negative half-cycle is the other half where ${\displaystyle r\leq 0}$.)
  • The shape of each petal is same because the graphs of half-cycles have the same shape. The shape is given by the positive half-cycle with crest at ${\displaystyle (a,0)}$ specified by ${\displaystyle r=a\cos(k\theta )}$ (that is bounded by the angle interval ${\displaystyle -T/4\leq \theta \leq T/4}$). The petal is symmetric about the polar axis. All other petals are rotations of this petal about the pole, including those for roses specified by the sine function with same values for ${\displaystyle a}$ and ${\displaystyle k}$.
  • Consistent with the rules for plotting points in polar coordinates, a point in a negative half-cycle cannot be plotted at its polar angle because its radial coordinate ${\displaystyle r}$ is negative. The point is plotted by adding ${\displaystyle \pi }$ radians to the polar angle with a radial coordinate ${\displaystyle |r|}$. Thus, positive and negative half-cycles can be coincident in the graph of a rose. In addition, roses are inscribed in the circle ${\displaystyle r=a}$.
  • When the period ${\displaystyle T}$of the sinusoid is less than or equal to ${\displaystyle 4\pi }$, the petal's shape is a single closed loop. A single loop is formed because the angle interval for a polar plot is ${\displaystyle 2\pi }$ and the angular width of the half-cycle is less than or equal to ${\displaystyle 2\pi }$. When ${\displaystyle T>4\pi }$ (or ${\displaystyle |k|\leq 1/2}$) the plot of a half-cycle can be seen as spiraling out from the pole in more than one circuit around the pole until plotting reaches the inscribed circle where it spirals back to the pole, intersecting itself and forming one or more loops along the way. Consequently, each petal forms 2 loops when ${\displaystyle 4\pi <T\leq 8\pi }$ (or ${\displaystyle 1/4\leq |k|<1/2}$), 3 loops when ${\displaystyle 8\pi <T\leq 12\pi }$ (or ${\displaystyle 1/6\leq |k|<1/4}$), etc.
  • A rose's petals will not intersect each other when the angular frequency ${\displaystyle k}$ is a non-zero integer; otherwise, petals intersect one another.

The rose ${\displaystyle r=\cos(4\theta )}$. Since ${\displaystyle k=4}$ is an even number, the rose has ${\displaystyle 2k=8}$ petals. Line segments connecting successive peaks lie on the circle ${\displaystyle r=1}$ will form an octagon. Since one peak is at ${\displaystyle (1,0)}$ the octagon makes sketching the graph relatively easy after the half-cycle boundaries are drawn.

When ${\displaystyle k}$ is a non-zero integer, the curve will be rose-shaped with ${\displaystyle 2k}$ petals if ${\displaystyle k}$ is even, and ${\displaystyle k}$ petals when ${\displaystyle k}$ is odd.[5]

• The rose is inscribed in the circle ${\displaystyle r=a}$, corresponding to the radial coordinate of all of its peaks.

• Because a polar coordinate plot is limited to polar angles between ${\displaystyle 0}$ and ${\displaystyle 2\pi }$, there are ${\displaystyle 2\pi /T=k}$ cycles displayed in the graph. No additional points need be plotted because the radial coordinate at ${\displaystyle \theta =0}$ is the same value at ${\displaystyle \theta =2\pi }$ (which are crests for two different positive half-cycles for roses specified by the cosine function).

• When ${\displaystyle k}$ is even (and non-zero), each of the ${\displaystyle 2k}$ half-cycles is graphed. Each peak corresponds to a point lying on the circle ${\displaystyle r=a}$. Line segments connecting successive peaks will form a regular polygon with an even number of vertices, and likewise, the roses are symmetric about the pole. The rose will be graphed in any continuous interval of polar angles ${\displaystyle 2\pi }$ radians long.

• When ${\displaystyle k}$ is odd, each of the ${\displaystyle k}$ positive half-cycles (or ${\displaystyle k}$ negative half-cycles) is graphed. Each peak corresponds to a point lying on the circle ${\displaystyle r=a}$. Each of curve’s negative half-cycles is “hidden” in that they are coincident with a positive half-cycle that will be displayed. This means that in graphing these rose curves, only the positive half-cycles or only the negative half-cycles need to plotted in order to form the full curve. (Equivalently, a complete curve will be graphed by plotting any continuous interval of polar angles that is ${\displaystyle \pi }$ radians long such as ${\displaystyle \theta =0}$ to ${\displaystyle \theta =\pi }$.[6]) Line segments connecting successive peaks will form a regular polygon with an odd number of vertices. The rose will be graphed in any continuous interval of polar angles ${\displaystyle \pi }$ radians long.

• The rose’s petals do not overlap.

A rose with ${\displaystyle k=1}$ is a circle with a diameter that lies on the polar axis when ${\displaystyle r=a\cos(\theta )}$. The circle is the curve's single petal. (See the circle being formed at the end of the next section.)

A rose with ${\displaystyle k=2}$ is called a quadrifolium because it has 4 petals.

A rose with ${\displaystyle k=3}$ is called a trifolium[7] because it has 3 petals. The curve is also called the Paquerette de Mélibée.[8] (See the trifolium being formed at the end of the next section.)

In general, when ${\displaystyle k}$ is a rational number in the form ${\displaystyle k=n/d}$, where ${\displaystyle n}$ and ${\displaystyle d}$ are non-zero integers, the number of petals is the denominator of the expression ${\displaystyle 1/2-1/(2k)=(n-d)/2n}$.[9] This means that the number of petals is ${\displaystyle n}$ if both ${\displaystyle n}$ and ${\displaystyle d}$ are odd, and ${\displaystyle 2n}$ otherwise.[10]

• In the case when both ${\displaystyle n}$ and ${\displaystyle d}$ are odd, the positive and negative half-cycles of the sinusoid are coincident. In all other cases the positive and negative half-cycles are not coincident. Consequently, if graphing is started at ${\displaystyle \theta =0}$, the full curve is completed at ${\displaystyle \theta =d\pi }$ with ${\displaystyle n}$ petals whenever both ${\displaystyle n}$ and ${\displaystyle d}$ are odd, and ${\displaystyle \theta =2d\pi }$ with ${\displaystyle 2n}$ petals otherwise. In the latter case, the roses are symmetric about the pole, and the roses specified by the cosine and sine functions are coincident.

• The rose is inscribed in the circle ${\displaystyle r=a}$, corresponding to the radial coordinate of all of its peaks.

• When ${\displaystyle d=1}$, ${\displaystyle k}$ is a non-zero integer, and we have the case described in the section immediately above. (When ${\displaystyle k}$ is an integer, ${\displaystyle n=k}$ and ${\displaystyle d=1}$, the number of petals is ${\displaystyle n=k}$ when ${\displaystyle k}$ is odd because ${\displaystyle n}$ and ${\displaystyle n=k}$ are odd, and ${\displaystyle 2n=2k}$ when ${\displaystyle k}$ is even.)

• When ${\displaystyle |k|\geq 1/2}$ each petal is a single closed loop; otherwise each petal has multiple loops because the period of the sinusoid has ${\displaystyle T>4\pi }$.

A rose with ${\displaystyle k=1/2}$ is called the Dürer folium, named after the German painter and engraver.[11] The roses specified by ${\displaystyle r=a\cos(\theta /2)}$ and ${\displaystyle r=a\sin(\theta /2)}$ are exactly the same even though ${\displaystyle a\cos(\theta /2)\neq a\sin(\theta /2)}$.

A rose with ${\displaystyle k=1/3}$ is a limaçon trisectrix which has the angle trisection property. The rose has a single petal with two loops. (See the animation below.)

Examples of roses ${\displaystyle r=\cos(k\theta )}$ created using gears with different ratios.
The rays displayed are the polar axis and ${\displaystyle \theta =\pi /2}$.
Graphing starts at ${\displaystyle \theta =0}$.

The circle, k=1 (n=1, d=1). The rose is complete when ${\displaystyle \theta =\pi }$ is reached (one-half revolution of the lighter gear).

The limaçon trisectrix, k=1/3 (n=1, d=3), has one petal with two loops. The rose is complete when ${\displaystyle \theta =3\pi }$ is reached (one and one-half revolution of the lighter gear).

The trifolium, k=3 (n=3, d=1). The rose is complete when ${\displaystyle \theta =\pi }$ is reached (one-half revolution of the lighter gear).

The 8 petals of the rose with k=4/5 (n=4, d=5) is each, a single loop that intersect other petals. The rose is symmetric about the pole. The rose is complete at ${\displaystyle \theta =10\pi }$ (five revolutions of the lighter gear).

A rose curve specified with an irrational number for ${\displaystyle k}$ will never complete. Furthermore, the graph of the curve in this case forms a dense set (i.e., it comes arbitrarily close to every point in the unit disk).