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added plot for cubic functions; some restructuring
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Martin Thoma
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2 changed files with 111 additions and 46 deletions
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@ -11,25 +11,22 @@ def euclideanDist(p1, p2):
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from math import sqrt
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return sqrt((p1.x-p2.x)**2 + (p1.y-p2.y)**2)
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def getMinDist(y, precision=0.001, startX=0, endX=3):
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"""Get x of point on (x,x^2) that has minimal distance to given Point."""
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def getMinDist(p1, precision=0.001, startX=0, endX=3):
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"""Get x of point on (x,x^2) that has minimal distance to given Point p."""
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minDist = -1
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p1 = Point(0, y)
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for x in numpy.arange(startX, endX, precision):
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p2 = Point(x, x**2)
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dist = euclideanDist(p1, p2)
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if minDist == -1 or dist < minDist:
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minDist = dist
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#print(dist)
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else:
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if abs(i-minDist) <0.005:
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print("x: %s" % str(x))
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break
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return minDist
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for i in numpy.arange(0, 3, 0.01):
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minDist = getMinDist(i)
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"""for i in numpy.arange(0, 3, 0.01):
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minDist = getMinDist(Point(0, i))
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if abs(i-minDist) < 0.005:
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print(i, minDist)
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print(i, minDist)"""
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print(getMinDist(Point(0,4.25), precision=0.001, startX=0, endX=3))
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#print(euclideanDist(Point(0,5),Point(2, 2**2)))
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#print(getMinDist(5, 0.00001, 2, 3))
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@ -86,15 +86,23 @@ But minimizing $d_{P,f}$ is the same as minimizing $d_{P,f}^2$:
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&= x_p^2 - 2x_p x + x^2 + y_p^2 - 2y_p f(x) + f(x)^2
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\end{align}
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\todo[inline]{hat dieser Satz einen Namen? kann ich den irgendwo zitieren? Ich habe ihn aus \href{https://github.com/MartinThoma/LaTeX-examples/tree/master/documents/Analysis I}{meinem Analysis I Skript} (Satz 22.3).}
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\begin{theorem}[Minima of polynomial functions]\label{thm:minima-of-polynomials}
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Let $n \in \mathbb{N}, n \geq 2$, $f$ polynomial function of
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degree $n$, $x_0 \in \mathbb{R}$, \\
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$f'(x_0) = f''(x_0) = \dots = f^{(n-1)} (x_0) = 0$
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and $f^{(n)} > 0$.
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\todo[inline]{Hat dieser Satz einen Namen? Gibt es ein gutes Buch,
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aus dem ich den zitieren kann? Ich habe ihn aus \href{https://github.com/MartinThoma/LaTeX-examples/tree/master/documents/Analysis I}{meinem Analysis I Skript} (Satz 21.5).}
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\begin{theorem}\label{thm:required-extremum-property}
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Let $x_0$ be a relative extremum of $f$.
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Then $x_0$ is a local minimum of $f$.
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Then: $f'(x_0) = 0$.
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\end{theorem}
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%bzw. 22.3
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%\begin{theorem}[Minima of polynomial functions]\label{thm:minima-of-polynomials}
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% Let $n \in \mathbb{N}, n \geq 2$, $f$ polynomial function of
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% degree $n$, $x_0 \in \mathbb{R}$, \\
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% $f'(x_0) = f''(x_0) = \dots = f^{(n-1)} (x_0) = 0$
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% and $f^{(n)} > 0$.
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%
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% Then $x_0$ is a local minimum of $f$.
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%\end{theorem}
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\clearpage
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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@ -251,28 +259,36 @@ $b, c \in \mdr$ be a quadratic function.
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\end{figure}
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\subsection{Calculate points with minimal distance}
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We use Theorem~\ref{thm:minima-of-polynomials}:\nobreak
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In this case, $d_{P,f}^2$ is polynomial of degree 4.
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We use Theorem~\ref{thm:required-extremum-property}:\nobreak
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\begin{align}
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0 &\stackrel{!}{=} (d_{P,f}^2)'\\
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&= -2 x_p + 2x -2y_p f'(x) + \left (f(x)^2 \right )'\label{eq:minimizingFirstDerivative}\\
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0 &\overset{!}{=} (d_{P,f}^2)'\\
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&= -2 x_p + 2x -2y_p f'(x) + \left (f(x)^2 \right )'\\
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&= -2 x_p + 2x -2y_p f'(x) + 2 f(x) \cdot f'(x) \rlap{\hspace*{3em}(chain rule)}\label{eq:minimizingFirstDerivative}\\
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&= -2 x_p + 2x -2y_p (2ax+b) + ((ax^2+bx+c)^2)'\\
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&= -2 x_p + 2x -2y_p \cdot 2ax-2 y_p b + (a^2 x^4+2 a b x^3+2 a c x^2+b^2 x^2+2 b c x+c^2)'\\
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&= -2 x_p + 2x -4y_p ax-2 y_p b + (4a^2 x^3 + 6 ab x^2 + 4acx + 2b^2 x + 2bc)\\
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&= 4a^2 x^3 + 6 ab x^2 + 2(1 -2y_p a+ 2ac + b^2)x +2(bc-by_p-x_p)\\
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0 &\stackrel{!}{=}(d_{P,f}^2)''\\
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&= 12a^2 x^2 + 12abx + 2(1 -2y_p a+ 2ac + b^2)
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&= 4a^2 x^3 + 6 ab x^2 + 2(1 -2y_p a+ 2ac + b^2)x +2(bc-by_p-x_p)
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\end{align}
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%\begin{align}
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% 0 &\overset{!}{=}(d_{P,f}^2)''\\
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% &= 2 - 2y_p f''(x) + \left ( 2 f(x) \cdot f'(x) \right )' \rlap{\hspace*{3em}(Eq. \ref{eq:minimizingFirstDerivative})}\\
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% &= 2 - 2y_p f''(x) + 2 \cdot \left ( f'(x) \cdot f'(x) + f(x) \cdot f''(x) \right ) \rlap{\hspace*{3em}(product rule)}\\
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% &= 2 - 2y_p f''(x) + 2 \cdot \left ( f'(x)^2 + f(x) \cdot f''(x) \right )\\
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% &= 12a^2 x^2 + 12abx + 2(1 -2y_p a+ 2ac + b^2)
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%\end{align}
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This is an algebraic equation of degree 3.
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There can be up to 3 solutions in such an equation. An example is
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There can be up to 3 solutions in such an equation. Those solutions
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can be found with a closed formula.
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\begin{align*}
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a &= 1 & b &= 0 & c &= 1 & x_p &= 0 & y_p &= 1
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\end{align*}
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\todo[inline]{Where are those closed formulas?}
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So $f(x) = x^2 + 1$ and $P(0, 1)$.
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\begin{example}
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Let $a = 1, b = 0, c= 1, x_p= 0, y_p = 1$.
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So $f(x) = x^2 + 1$ and $P(0, 1)$.
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\begin{align}
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0 &\stackrel{!}{=} 4 x^3 - 2x\\
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@ -281,6 +297,7 @@ So $f(x) = x^2 + 1$ and $P(0, 1)$.
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\end{align}
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As you can easily verify, only $x_1$ is a minimum of $d_{P,f}$.
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\end{example}
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\subsection{Number of points with minimal distance}
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@ -288,10 +305,18 @@ It is obvious that a quadratic function can have two points with
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minimal distance.
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For example, let $f(x) = x^2$ and $P = (0,5)$. Then $P_{f,1} \approx (2.179, 2.179^2)$
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has minimal distance to $P$, but also $P_{f,2}\approx (-2.179, 2.179^2)$.
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has minimal distance to $P$, but also $P_{f,2}\approx (-2.179, 2.179^2)$.\todo{exact example?}
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Obviously, there cannot be more than three points with minimal distance.
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But can there be three points?
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As discussed before, there cannot be more than 3 points on the graph
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of $f$ next to $P$.
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\todo[inline]{But can there be three points? O.b.d.A: $a > 0$.
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As $c$ is moves the curve only up and down, we can o.b.d.A assume
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that $c=0$.
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$x=-\frac{b}{2a}$ is the minimum of $f$. If there are 3 points, this will
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be one of them. The other two ones are symmetric by an axis through
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$-\frac{b}{2a}$}
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\begin{figure}[htp]
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\centering
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@ -328,7 +353,7 @@ But can there be three points?
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\caption{3 points with minimal distance?}
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\end{figure}
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\todo[inline]{write this}
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\clearpage
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% Cubic %
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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@ -337,29 +362,72 @@ Let $f(x) = a \cdot x^3 + b \cdot x^2 + c \cdot x + d$ be a cubic function
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with $a \in \mdr \setminus \Set{0}$ and
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$b, c, d \in \mdr$ be a function.
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\subsection{Number of points with minimal distance}
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\begin{figure}[htp]
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\centering
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\begin{tikzpicture}
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\begin{axis}[
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legend pos=south east,
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axis x line=middle,
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axis y line=middle,
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grid = major,
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width=0.8\linewidth,
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height=8cm,
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grid style={dashed, gray!30},
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xmin=-3, % start the diagram at this x-coordinate
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xmax= 3, % end the diagram at this x-coordinate
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ymin=-3, % start the diagram at this y-coordinate
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ymax= 3, % end the diagram at this y-coordinate
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axis background/.style={fill=white},
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xlabel=$x$,
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ylabel=$y$,
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%xticklabels={-2,-1.6,...,7},
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%yticklabels={-8,-7,...,8},
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tick align=outside,
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minor tick num=-3,
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enlargelimits=true,
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tension=0.08]
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\addplot[domain=-3:3, thick,samples=50, red] {x*x*x};
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\addplot[domain=-3:3, thick,samples=50, green] {x*x*x+x*x};
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\addplot[domain=-3:3, thick,samples=50, blue] {x*x*x+2*x*x};
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\addplot[domain=-3:3, thick,samples=50, orange] {x*x*x+x};
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\addlegendentry{$f_1(x)=x^3$}
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\addlegendentry{$f_2(x)=x^3 + x^2$}
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\addlegendentry{$f_2(x)=x^3 + 2 \cdot x^2$}
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\addlegendentry{$f_1(x)=x^3 + x$}
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\end{axis}
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\end{tikzpicture}
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\caption{Cubic functions}
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\end{figure}
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\todo[inline]{Write this}
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%
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%\subsection{Special points}
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%\todo[inline]{Write this}
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%
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%\subsection{Voronoi}
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%
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%For $b^2 \geq 3ac$
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%
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%\todo[inline]{Write this}
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\subsection{Special points}
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\todo[inline]{Write this}
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\subsection{Voronoi}
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For $b^2 \geq 3ac$
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\todo[inline]{Write this}
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\subsection{Calculate points with minimal distance}
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When you want to calculate points with minimal distance, you can
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take the same approach as in Equation \ref{eq:minimizingFirstDerivative}:
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\begin{align}
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0 &\stackrel{!}{=} -2 x_p + 2x -2y_p(f(x))' + (f(x)^2)'\\
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\Leftrightarrow 0 &\stackrel{!}{=} 2 f(x) \cdot f'(x) - 2 y_p f'(x) + 2x - 2 x_p\\
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\Leftrightarrow 0 &\stackrel{!}{=} \underbrace{\left (2 f(x) - 2 y_p \right ) \cdot f'(x)}_{\text{Polynomial of degree 5}} + \underbrace{2x - 2 x_p}_{\text{:-(}}
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&= 2 f(x) \cdot f'(x) - 2 y_p f'(x) + 2x - 2 x_p\\
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&= \underbrace{\left (2 f(x) - 2 y_p \right ) \cdot f'(x)}_{\text{Polynomial of degree 5}} + \underbrace{2x - 2 x_p}_{\text{:-(}}
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\end{align}
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\subsection{Number of points with minimal distance}
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As there is an algebraic equation of degree 5, there cannot be more
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than 5 solutions.
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\todo[inline]{Can there be 3, 4 or even 5 solutions? Examples!
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After looking at function graphs of cubic functions, I'm pretty
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sure that there cannot be 4 or 5 solutions, no matter how you
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chose the cubic function $f$ and $P$.
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\todo[inline]{Write this}
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I'm also pretty sure that there is no polynomial (no matter what degree)
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that has more than 3 solutions.}
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\end{document}
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