45°旋转对称的函数

xcx

定义在 $R$ 上函数 $f$ 满足\[ \frac{x+f(x)}{\sqrt2}=f(\frac{x-f(x)}{\sqrt2}) \]
若存在非零实数 $x_0$ 使\[ f(x_0) \sin \theta=f(f(x_0))\cos \theta \]求 $\theta$ 取值范围

hbghlyj

$(x,y)$在$f$的图象上,则$\left(\frac{x-y}{\sqrt2},\frac{x+y}{\sqrt2}\right)$在$f$的图象上.所以$f$的图象关于原点有45°的旋转对称.

例如:以原点为中心,(1,0)为一个顶点作正八边形,将线段(1,0)--$\big(\frac{1+\cos\fracπ8}2,\frac{\sin\fracπ8}2\big)$绕原点旋转$\frac{nπ}8(n=0,1,⋯,7)$,这八条线段构成一个函数图象(端点处适当选择包含),容易看出它是关于原点45°旋转对称的.
反思:
例子不正确⋯⋯端点处的实心和空心有区别,无法做到旋转45°对称性

kuing

回复 2# hbghlyj

有点儿意思
PS、我优化了一下你那图的代码，顺便加上了空心圈。

hbghlyj

回复 4# kuing

假如函数上有一点为$(0,f(0))$,它旋转180°以后到了$(0,-f(0))$,不能有两个点的横坐标相等,所以$f(0)=0$.

假如函数上有一点为$(a,a),a≠0$,它旋转-90°以后到了$(a,-a)$,不能有两个点的横坐标相等,矛盾.所以函数图象与$y=x(x≠0)$没有公共点.同理函数图象与$y=-x(x≠0)$没有公共点.

回到原题,说的是存在$x_0$使$f(x_0) \sin \theta=f(f(x_0))\cos \theta$,但是因为$f$是定义在$\Bbb R$上的,而且是90°旋转对称的,所以$f$的值域是$\Bbb R$,所以对每个$x∈\Bbb R$都存在$x_0$使$f(x_0)=x$.
所以说的就是存在$x$使$x\sin \theta=f(x)\cos \theta$,也就是$\tan\theta=\frac{f(x)}{x}$,所以θ是$(y,f(x))$与x轴正半轴的夹角.

hbghlyj

math.stackexchange.com/questions/1391011/what … f-order-other-than-2
Below are the examples of graphs with 4-fold symmetry. The functions are piecewise continuous (and even piecewise linear, though this could easily have been done otherwise) and defined on entire $\Bbb R$.

As you may see, the first function doesn't have any degrees of freedom (except scaling, which is not interesting), but the other one does (the mentioned a). One may wonder whether it is possible to accommodate for the six-fold thing by sacrificing that degree of freedom. Now, lo and behold...

The slopes of the linear segments are (top to bottom) $\sqrt5+2\over\sqrt3$,  $\sqrt5-2\over\sqrt3$, and $-{\sqrt5\over\sqrt3}$. The points of discontinuity form a geometric progression, whose common ratio (as luck would have it) is the golden ratio.



如何构造出8-fold symmetry的函数呢

hbghlyj

如何构造出 8-fold symmetry 的函数呢

ljh25252

1. Plot[{(-1)^(Ceiling[Abs[x]] + 1) x + Sign[x]}, {x, -10, 10},
2. PlotRange -> Full, Frame -> True, AspectRatio -> 1,
3. ExclusionsStyle -> {Directive[Dashing[Small], Red], Red}]

Copy the Code

$$
(-1)^{\lceil\lvert x\rvert\rceil+1}x+\mathrm{sgn}x
$$


$$
\beta=2\arctan\left[\frac{1}{3}\left(2\sqrt{3(3+\sqrt{5})}-2\sqrt{3}-\sqrt{15}\right)\right]
$$

kuing

刚才和 585 又聊起这题，用 tikz 画了个图（5# 的右图）：

往里面再多画一段，只要稍微改下代码（foreach 的好处），顺便加点颜色：