I need to create a polynomial function that passes through the points $(0,0)$, $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ where $$ x_1 < x_2 < x_3\quad \text{ and } \quad y_1 < y_2 < y_3 $$ with the only constraint that that function has to be increasing for all $x > 0$. I don't mind the degree of the polynomial function.

I am using now a spline approach to build such a function, but I am trying to substitute by a polynomial. The Lagrange interpolation does not ensure that the function is increasing.

  • $\begingroup$ But Lagrange ensures that the function is always increasing? $\endgroup$ – tonecho Nov 12 '14 at 12:27
  • $\begingroup$ It's a hard problem if you want a single polynomial. Begin by looking up "monotonicity preserving polynomial interpolation", but I don't know of a definitive source for this. Typically one uses splines for this: see Monotone cubic interpolation $\endgroup$ – user147263 Nov 12 '14 at 12:50
  • $\begingroup$ I am using now a spline approach to build a function, but I am trying to substitute by a polynom. $\endgroup$ – tonecho Nov 13 '14 at 7:33
  • $\begingroup$ @tonecho (Most) splines are polynomials - see, for instance, en.wikipedia.org/wiki/B%C3%A9zier_curve#Polynomial_form , which both gives an explicit polynomial form and also gives several excellent reasons not to use it. $\endgroup$ – Steven Stadnicki Jan 7 '16 at 23:56
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    $\begingroup$ @StevenStadnicki I hold the view that piecewise linear functions are not linear functions, and similarly, that polynomial splines are not polynomials... $\endgroup$ – user147263 Jan 8 '16 at 22:49

Here are som thoughts abouth this question, but they do not lead to a conclusive statement.

I will use the following evidence:

If a polynomial $P$ is such that $P(a)=b$, then $P$ is necessarily of the form $$ P(X)=(X-a)Q(X)+b,$$ where $Q$ is a polynomial.

The idea is to apply this iteratively to $P$ such that $P(0)=0$, $P(x_i)=y_i$ with $1\leq i\leq3$. Since $P(0)=0$, $P$ is of the form $$P=XA(X)$$ where $A$ is a polynomial. Let us compute $A(x_1)$. We have $P(x_1)=x_1A(x_1)=y_1$, so we have $$ A(x_1)=\frac{y_1}{x_1}\equiv z_1.$$ We conclude that $A$ is of the form $$A(X)=z_1+(X-x_1)B(X)$$ where $B$ is a polynomial. We can compute $B(x_2)$ using $$y_2=P(x_2)=x_2A(x_2)=x_2\big[z_1+(x_2-x_1)B(x_2)\big].$$ We get $$B(x_2)=\frac{1}{x_2-x_1}\left(\frac{y_2}{x_2}-\frac{y_1}{x_1}\right)\equiv z_2.$$ We can likewise write $B$ as $$B(X)=z_2+(X-x_2)C(X),$$ where $C$ is a polynomial. We compute $C(x_3)$ using $$\begin{split}y_3&=P(x_3)=x_3A(x_3)=x_3\Big[z_1+(x_3-x_1)B(x_3)\Big]\\&=x_3\Bigg\{z_1+(x_3-x_1)\Big[z_2+(x_3-x_2)C(x_3)\Big]\Bigg\}.\end{split}$$ We obtain $$C(x_3)=\frac{1}{x_3-x_2} \left[\frac{1}{x_3-x_1}\left(\frac{y_3}{x_3}-\frac{y_2}{x_1}\right)-z_2\right]\equiv z_3.$$ Finally (or at last) we can write $C$ as $$C(X)=z_3+(X-x_3)Q(X).$$ By necessary conditions, any polynomial satisfying the requirements $P(0)=0$, $P(x_i)=y_i$ is of the form $$P(X)=X\Bigg\{z_1+(X-x_1)\Big[z_2+(X-x_2)\big(z_3+(X-x_3)Q(X)\big)\Big]\Bigg\}.$$

Clearly if $Q=0$ then we have the Lagrange interpolation result. If this polynomial does not work, then we must consider $Q\neq0$ with a positive leading coefficient.

We must have $P'(0)\geq0$. Let us compute $P'(0)$. We have $$P'(0)=A(0)=z_1-x_1z_2+x_1x_2z_3-x_1x_2x_3Q(0).$$ Consequently, we want $$Q(0)\leq\frac{z_1-x_1z_2+x_1x_2z_3}{x_1x_2x_3}\equiv q_0.$$ If $q_0\leq0$ then necessarily $Q$ must be of degree at least 1. By considering $P'(x_i)\geq0$, we can find an inequality for $Q(x_i)$. This sums up to four inequalities, so the required minimal degree of $Q$ for the general case is probably at least $3$, unless I'm missing something. I believe the general case requires a computer's help.

This does not provide a full answer, I'm afraid. I anyway hope this could help.


The task permits full solution in case of points $$(0,0), (x_1,y_1), (x_2,y_2),\quad 0<x_1<x_2,\quad 0<y_1<y_2.$$
Let $$t=\dfrac{x}{x_2},\quad f=\dfrac{y}{y_2},$$ then points choosing to $$(0,0), (t,f), (1,1),\quad t,f\in[0,1].$$ Let consider polynomial function $$y(x)=ax+bx^2+cx^3,$$ $$y'(x)=a+2bx+3bx^2,\quad y'(x)>0, \text{ if }x\in (0,1).$$
Known that minimal value of derivation achieves on segment bounds or in inflection point if it belongs to segment $(0,1)$. Coordinates of this point can be calculated from equation $y''(t_i) = 0, \quad 2b+3bt_i = 0$ or $t_i = -\dfrac b{3c}$, so

$$\begin{cases} at + bt^2 + ct^3 = f\\ a + b + c = 1\\ a > 0\\ a + 2b + 3c > 0\\ a - \dfrac{b^2}{3c} > 0, \text{ if }-\dfrac {b}{3c}\in(0,1), \end{cases}$$ $$\begin{cases} a + b + c = 1\\ b(t-t^2) + c(t-t^3) = t-f\\ b + c < 1\\ b + 2c > -1\\ b + c + \dfrac{b^2}{3c} < 1, \text{ if } bc\in(-3c^2,0) \end{cases}.$$

The last inequality can be written as $$ \left[ \genfrac{.}{.}{0pt}{0} {\begin{cases} c < 0\\ b \in (0, -3c)\\ b^2+3bc+3c^2-3c > 0 \end{cases}} {\begin{cases} c > 0\\ b\in(-3c,0)\\ b^2+3bc+3c^2-3c < 0 \end{cases}} \right.$$ Discriminant of polinomial $b^2+3bc+3c^2-3c$ (when $c$ regarded as a parameter) is $$d = (3c)^2 - 4(3c^2-3c) = 12c-3c^2.$$

$d<0$ when $c\not\in[0,4],$ so: $$\left[ \genfrac{.}{.}{0pt}{0} {\begin{cases} c < 0\\ b \in (0, -3c)\\ \end{cases}} {\begin{cases} c\in[0,4]\\ b\in[b_1,b_2]\\ \end{cases}} \right. $$ where $$b_1 = \max(-3c, -1.5c-\sqrt{3c-0.75c^2},$$ $$b_2 = \min(0, -1.5c+\sqrt{3c-0.75c^2}.$$ Built using Wolfram Alpha (1) graph shows that when $c\geq0$ this restriction It does not break the connection area of the allowed parameter values: graph When $c<0,$ similar graph shows us allowed zone as two triangles with vertices in the points $(0,0), (0,1), (1.5,0.5)$ and $(0,-1),(0,0),(-0.5,0):$ enter image description here

General picture shows us allowed zone as two triangles with vertices in the points $(0,0), (0,1), (1.5,0.5)$ and $(0,-1),(0,0),(-0.5,0):$ general

If do not take into account the limitations of $c\in[0,4]$, the shape of the permitted area resembles forked spear.

Thus $$\begin{cases} a + b + c = 1\\ b = \dfrac{t-f}{t-t^2} - c(1+t)\\ b + c < 1\\ b + 2c > -1\\ \left[ \genfrac{.}{.}{0pt}{0} {\begin{cases} c < 0\\ b \in (0, -3c)\\ \end{cases}} {\begin{cases} c\in[0,4]\\ b\in[b_1,b_2]\\ \end{cases}} \right. \end{cases}.$$

Area restrictions defined, and it remains to determine whether to always find common points of the second equation of the system with this area. Since the second equation is infinite straight line, the only reason for the lack of a decision may be the output of the slope of this line beyond (-2,-1) angular coefficients of the second and third inequalities.

The form of the equation that this requirement is satisfied. Consequently, in the problem there are always values $b,c$ on the allowed area and $a=1-b-c$ in which the problem in this formulation has a solution.


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