# Finding a monotone function crossing three given points

• Jun 30th 2008, 02:01 PM
Sogartar
Finding a monotone function crossing three given points
Hi,
I have been trying from quite a while to solve the following problem.
Find a real differentiable monotone function $f(x)$ in the interval $[0, c]$, such that:
$f(0) = 0$
$f(b) = 1/2$
$f(c) = 1$,
where $0 < b < c$.
This is a particular case of more general problem:
Find a real differentiable monotone function $f(x)$ in the interval $[a, c]$, such that:
$f(a) = A$
$f(b) = B$
$f(c) = C$,
where $a < b < c$ and $A < B < C$.
I tried several different ways but ended up with some bad looking equations. My method was to define f as a function of some type with three parameters and try find these parameters from the above equations.
Example, $f(x) = pe^{qx} + r$, where $p$, $q$ and $r$ are real parameters.
If applied to the more particular first case, I end up with the equation
$2e^{bq} - e^{cq} - 1 = 0$.
I would be very grateful if you know such a function and share it.
• Jun 30th 2008, 02:04 PM
Mathstud28
Quote:

Originally Posted by Sogartar
Hi,
I have been trying from quite a while to solve the following problem.
Find a real differential monotone function f(x) in the interval [0, c], such that:
f(0) = 0
f(b) = 1/2
f(c) = 1,
where 0 < b < c.
This is a particular case of more general problem:
Find a real monotone function f(x) in the interval [a, c], such that:
f(a) = A
f(b) = B
f(c) = C,
where a < b < c and A < B < C.
I tried several different ways but ended up with some bad looking equations. My method was to define f as a function of some type with three parameters and try find these parameters from the above equations.
Example, f(x) = p*Exp(qx) + r, where p, q and r are real parameters.
If applied to the more particualr first case, I end up with the equation
2*Exp(bq) - Exp(cq) - 1 = 0.
I would be very grateful if know such a function and share it.

Maybe I am misunderstanding you but

Let $f(x)=x$

We see that $f(x)$ is monotonically increasing $\forall{x}\in\mathbb{R}$

We also see that if we let $b=\frac{1}{2}$ and $c=1$

We have that

$[0,1]$

Where $0<\frac{1}{2}<1$

and we also have that

$f(0)=0$

and

$f\left(\frac{1}{2}\right)=\frac{1}{2}$

and

$f(1)=1$
• Jun 30th 2008, 04:21 PM
algebraic topology
Quote:

Originally Posted by Mathstud28
We also see that if we let $b=\frac{1}{2}$ and $c=1$

No, b and c are given. You can’t choose your own values for them.
• Jul 1st 2008, 04:26 AM
Sogartar
Solution
Thanks for your consideration on my problem.
Today while thinking on it, I decided to apply the old Roman tactic, divide and conquer. So my plan was to find two functions. The first in the interval $[0, b]$, crossing $(0,0)$ and $(b,1/2)$. The second in the interval $(b, c]$, crossing $(b,1/2)$ and $(c, 1)$. Also I needed those functions to have equal derivatives in point $b$.
So, my first guess was two parabolas $F(x)\in[0, b]$ and $G(x)\in(b, c]$, where $F'(b) = G'(b) = 0$. This is a solution of the problem at hand, but the characteristics of this function were not what I was looking for. I needed changes in $x$ around $b$ to yell greater changes in $f(x)$, than around the ends of the interval $[0, c]$. So I defined $F(x)$ and $G(x)$ such that, $F^{-1}(x)$ and $G^{-1}(x)$ to be parabolas with the above requirements.
Another solution was to use ellipses. Taking the correct quarter of each ellipse where, the first one is with center $(0, 1/2)$ and passes through $(0,0)$ and $(b,1/2)$. The second is with center $(c,1/2)$ and passing through $(b, 1/2)$ and $(c, 1)$. This defines a nice monotone differentiable function in $(0, c)$.
• Jul 1st 2008, 04:55 AM
Isomorphism
Quote:

Originally Posted by Sogartar
Hi,
I have been trying from quite a while to solve the following problem.
Find a real differentiable monotone function $f(x)$ in the interval $[0, c]$, such that:
$f(0) = 0$
$f(b) = 1/2$
$f(c) = 1$,
where $0 < b < c$.

EDIT: The function I found is not necessarily monotone, so sorry.

Try defining such a function:

Let $f(x) = A(x-b)(x-c) + Bx(x-b)+Cx(x-c)$, where A, B, C are constants we will work out.

$f(0) = Abc = 0\Rightarrow A = 0$
$f(b) = Cb(b-c) = \frac12 \Rightarrow C = \frac{1}{2b(b-c)}$
$f(c) = Bc(c-b) = 1 \Rightarrow C = -\frac{1}{b(b-c)}$

I think this should be differentiable since its a polynomial function. All that remains is to restrict the function to the interval [0,c].
• Jul 1st 2008, 05:10 AM
CaptainBlack
Quote:

Originally Posted by Sogartar
Hi,
I have been trying from quite a while to solve the following problem.
Find a real differentiable monotone function $f(x)$ in the interval $[0, c]$, such that:
$f(0) = 0$
$f(b) = 1/2$
$f(c) = 1$,
where $0 < b < c$.
This is a particular case of more general problem:
Find a real differentiable monotone function $f(x)$ in the interval $[a, c]$, such that:
$f(a) = A$
$f(b) = B$
$f(c) = C$,
where $a < b < c$ and $A < B < C$.
I tried several different ways but ended up with some bad looking equations. My method was to define f as a function of some type with three parameters and try find these parameters from the above equations.
Example, $f(x) = pe^{qx} + r$, where $p$, $q$ and $r$ are real parameters.
If applied to the more particular first case, I end up with the equation
$2e^{bq} - e^{cq} - 1 = 0$.
I would be very grateful if you know such a function and share it.

Try finding constants: $c_1>0$, $c_2>0$, $c_3$, such that:

$f(x)=c_1e^{c_2 x}+c_3$

that satisfy the conditions.

RonL

ronL
• Jul 1st 2008, 08:16 AM
algebraic topology
Quote:

Originally Posted by Isomorphism
Try defining such a function:

Let $f(x) = A(x-b)(x-c) + Bx(x-b)+Cx(x-c)$, where A, B, C are constants we will work out.

$f(0) = Abc = 0\Rightarrow A = 0$
$f(b) = Cb(b-c) = \frac12 \Rightarrow C = \frac{1}{2b(b-c)}$
$f(c) = Bc(c-b) = 1 \Rightarrow C = -\frac{1}{b(b-c)}$

I think this should be differentiable since its a polynomial function. All that remains is to restrict the function to the interval [0,c].

It may not be monotone though.

I did think of CaptainBlack’s method, but I got some rather horrible-looking expressions for the coefficients and gave up.
• Jul 1st 2008, 08:26 AM
Isomorphism
Quote:

Originally Posted by algebraic topology
It may not be monotone though.

Whoops! My bad... I did not notice that (Punch)
• Jul 1st 2008, 02:34 PM
algebraic topology
Well, here’s a modification of CaptainBlack’s suggestion.

Join the first two points ( $(a,A)$ and $(b,B)$) by a straight line. This will have gradient $\frac{B-A}{b-a}$. Then join $(b,B)$ to $(c,C)$ by the function $y=f(x)=C_1\mathrm{e}^{C_2x}+C_3$, where $f(b)=B$, $f(c)=C$, and $f'(b)=\frac{B-A}{b-a}$.

Hopefully that might makes things a bit simpler.

By the way, CaptainBlack, the condition $C_1,C_2>0$ will make $f(x)$ convex, and this will only work if $(b,B)$ lies below the straight line joining $(a,A)$ and $(c,C)$. If $(b,B)$ lies above that line, then we need a concave function, which means we need both $C_1$ and $C_2$ to be negative ( $C_2<0$ to make the function concave and $C_1<0$ to make it increasing).
• Jul 1st 2008, 03:22 PM
CaptainBlack
Quote:

Originally Posted by algebraic topology
Well, here’s a modification of CaptainBlack’s suggestion.

Join the first two points ( $(a,A)$ and $(b,B)$) by a straight line. This will have gradient $\frac{B-A}{b-a}$. Then join $(b,B)$ to $(c,C)$ by the function $y=f(x)=C_1\mathrm{e}^{C_2x}+C_3$, where $f(b)=B$, $f(c)=C$, and $f'(b)=\frac{B-A}{b-a}$.

Hopefully that might makes things a bit simpler.

By the way, CaptainBlack, the condition $C_1,C_2>0$ will make $f(x)$ convex, and this will only work if $(b,B)$ lies below the straight line joining $(a,A)$ and $(c,C)$. If $(b,B)$ lies above that line, then we need a concave function, which means we need both $C_1$ and $C_2$ to be negative ( $C_2<0$ to make the function concave and $C_1<0$ to make it increasing).

I expect my mental image of this was convex, so it works for what I was trying to do, its just that what I was trying to do was only half the problem!

In fact we need no restrictions on the constants, with the given relations between the values at the given points a function of the given form which passes through the data will be monotone increasing.

RonL