# Given f(x) = x^3-3x, how can you construct an infinitely differentiable one-one function g(x):RR->RR with g(x) = f(x) in (-oo, -2] uu [2, oo)?

Oct 9, 2016

Here is my first attempt. (Unfortunately, it fails.)

#### Explanation:

$f \left(x\right) = {x}^{3} - 3 x$

graph{x^3-3x [-5.454, 5.645, -2.606, 2.94]}

My first thought is for a piecewise function, keeping the definition of $f \left(x\right)$ for $\left\mid x \right\mid \ge 2$ and inserting a piece on $\left[- 2 , 2\right]$ that matches at the joints, $- 2$ and $2$

I tried a piece based on $a {x}^{3}$, but could not get both the function and its derivative to agree at the joints.

We need $g \left(- 2\right) = - 2$ and $g \left(2\right) = 2$

and $g ' \left(- 2\right) = g ' \left(2\right) = 9$.

Determined to find a polynomial, I set

$g \left(x\right) = a {x}^{n}$ where $n$ is odd

So, $g ' \left(x\right) = n a {x}^{n - 1}$

We need $a {2}^{n} = 2$ and $n a {2}^{n - 1} = 9$

This lead to $n = 0$ and $a = \frac{1}{2} ^ 8$.

But for infinite differentiability, we also need

$g ' ' \left(2\right) = - 12$ and $g ' ' \left(2\right) = 12$

$g ' ' ' \left(- 2\right) = g ' ' ' \left(2\right) = 6$

${g}^{\left(n\right)} \left(- 2\right) = {g}^{\left(n\right)} \left(2\right) = 0$ for $n \ge 4$.

This attempt fails to satisfy those.

My initial thought was to try something using $\arcsin \left(\frac{x}{2}\right)$, but we need the fourth and subsequent derivatives to be $0$.

Perhaps we can start with and odd, four-times integrable function and work backwards from there.

So ${g}^{\left(4\right)} \left(x\right) = k \arcsin \left(\frac{x}{2}\right)$

Oct 9, 2016

$\cancel{\text{I am cautiously optimistic}}$. Never mind.

#### Explanation:

${g}^{\left(4\right)} \left(x\right) = \arcsin \left(\frac{x}{2}\right)$ is $0$ at the joints

This makes

$g ' ' ' \left(x\right) = \sqrt{4 - {x}^{2}} + x {\sin}^{-} 1 \left(\frac{x}{2}\right) + 6$

(All integrals by Wolfram Alpha) And

$g ' ' \left(x\right) = \frac{3}{4} x \left(\sqrt{4 - {x}^{2}} + 8\right) + \frac{1}{2} \left({x}^{2} + 2\right) {\sin}^{-} 1 \left(\frac{x}{2}\right)$>

(Note: $g ' ' \left(\pm 2\right) = \frac{3}{4} \left(\pm 2\right) \left(8\right) = \pm 12$)

$g ' \left(x\right) = \left({x}^{3} / 6 + x\right) {\sin}^{-} 1 \left(\frac{x}{2}\right) + \left(\frac{11 \sqrt{4 - {x}^{2}}}{36} + 3\right) {x}^{2} + \frac{4}{9} \left(\sqrt{4 - {x}^{2}} - 27\right) + 9$

Never mind.

What was I thinking? $g \left(5\right) \left(2\right)$ is not defined.

Oct 9, 2016

See below.

#### Explanation:

For practical purposes and with a guarantee of accuracy, we can proceed, according to

$g \left(x\right) = {\sum}_{k = 1}^{4} {a}_{k} {e}^{k x} + {a}_{k + 4} {e}^{- k x}$ and then given

${x}_{i} = - 2$ and ${x}_{s} = 2$ the conditions for ${a}_{k}$

{ (f = g), ((df)/(dx)=(dg)/(dx)), ((d^2f)/(dx^2)=(d^2g)/(dx^2)), ((d^3f)/(dx^3)=(d^3g)/(dx^3)) :}

This for ${x}_{i}$ and ${x}_{s}$ giving a set of $8$ linear equations in ${a}_{k} , k = 1 , 2 , \cdots , 8$

Solving this system we obtain

{ (a_1=-2.28226),(a_2=0.625881),(a_3=-0.0501427),(a_4=0.00155095),(a_5=2.28226),(a_6=-0.625881),(a_7=0.0501427),(a_8=-0.00155095) :}

So the binding function is ${\mathbb{C}}^{\infty}$

Attached a plot showing the binding.

Oct 9, 2016

$g \left(x\right) = \left\{\begin{matrix}{x}^{3} - 3 x & \text{ if " x in (-oo -2] uu [2 oo) \\ x^3-3x+e^(-x^2/(x^2-4)^2)*(x^9/256+3x-x^3)(1+x^2/(x^2-4)^2+x^4/(2(x^2-4)^4)) & " if } x \in \left(- 2 2\right)\end{matrix}\right.$

#### Explanation:

A classic example of a function for which the Taylor series fails is:

$j \left(x\right) = \left\{\begin{matrix}0 & \text{ if " x = 0 \\ e^(-1/x^2) & " if } x \ne 0\end{matrix}\right.$

This has the property that the function and all of its derivatives exist and are $0$ at $x = 0$. Hence the Taylor expansion at $x = 0$ would yield the zero function.

Inspired by this, consider the function:

$h \left(x\right) = \left\{\begin{matrix}0 & \text{ if " x = +-2 \\ e^(-x^2/(x^2-4)^2) & " otherwise}\end{matrix}\right.$

Then $h \left(x\right)$ and all of its derivatives exist and are $0$ when $x = \pm 2$. Additionally, $h \left(0\right) = {e}^{0} = 1$ ...

graph{e^(-x^2/(x^2-4)^2) [-5, 5, -2.5, 2.5]}

Given $f \left(x\right) = {x}^{3} - 3 x$

Define

$g \left(x\right) = \left\{\begin{matrix}f \left(x\right) & \text{ if " x in (-oo -2] uu [2 oo) \\ f(x)+h(x)(x^9/256-f(x))(1+x^2/(x^2-4)^2+x^4/(2(x^2-4)^4)) & " if } x \in \left(- 2 2\right)\end{matrix}\right.$

The idea here is that $g \left(x\right) = f \left(x\right)$ outside the interval $\left(- 2 , 2\right)$ and in the middle of the interval $\left(- 2 , 2\right)$, $g \left(x\right)$ behaves more like ${x}^{9} / 256$

graph{x^3-3x+e^(-x^2/(x^2-4)^2)*(x^9/256+3x-x^3)(1+x^2/(x^2-4)^2+x^4/(2(x^2-4)^4)) [-5, 5, -2.5, 2.5]}

The additional trick here is that the multiplier:

$\left(1 + {x}^{2} / {\left({x}^{2} - 4\right)}^{2} + {x}^{4} / \left(2 {\left({x}^{2} - 4\right)}^{4}\right)\right)$

is a sufficiently good approximation to ${e}^{{x}^{2} / {\left({x}^{2} - 4\right)}^{2}}$ without itself being an exponential function.

As a result, we do not lose the smoothing properties of $h \left(x\right)$ at the 'joins' at $x = \pm 2$

It remains to show that the resulting function is monotonically increasing.