Question

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)`?

Answer 1

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

Explanation:

`f(x) = x^3-3x`

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(x)` for `absx >= 2` and inserting a piece on `[-2,2]` that matches at the joints, `-2` and `2`

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

We need `g(-2)=-2` and `g(2) = 2`

and `g'(-2)=g'(2)=9`.

Determined to find a polynomial, I set

`g(x)=ax^n` where `n` is odd

So, `g'(x) = nax^(n-1)`

We need `a2^n=2` and `na2^(n-1)=9`

This lead to `n=0` and `a = 1/2^8`.

But for infinite differentiability, we also need

`g''(2) =-12` and `g''(2) = 12`

`g'''(-2) = g'''(2) = 6`

`g^((n))(-2) = g^((n))(2) = 0` for `n>= 4`.

This attempt fails to satisfy those.

My initial thought was to try something using `arcsin(x/2)`, 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^((4))(x) = k arcsin(x/2)`

Answer 2

`cancel("I am cautiously optimistic")`. Never mind.

Explanation:

`g^((4))(x) = arcsin(x/2)` is `0` at the joints

This makes

`g'''(x) = sqrt(4-x^2)+xsin^-1 (x/2) +6`

(All integrals by Wolfram Alpha) And

`g''(x) = 3/4 x(sqrt(4-x^2)+8)+1/2(x^2+2)sin^-1(x/2)`>

(Note: `g''(+-2) = 3/4(+-2)(8) = +-12`)

`g'(x) = (x^3/6 +x)sin^-1(x/2)+((11sqrt(4-x^2))/36+3)x^2+4/9(sqrt(4-x^2)-27)+9`

Never mind.

What was I thinking? `g(5)(2)` is not defined.

Answer 3

See below.

Explanation:

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

`g(x) = sum_(k=1)^4a_k e^(kx)+a_(k+4)e^(-kx)` 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 `CC^oo`

Attached a plot showing the binding.

Answer 4

`g(x) = { (x^3-3x, " 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 (-2, 2)) :}`

Explanation:

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

`j(x) = { (0, " if " x = 0), (e^(-1/x^2), " if " x != 0) :}`

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(x) = { (0, " if " x = +-2), (e^(-x^2/(x^2-4)^2), " otherwise") :}`

Then `h(x)` and all of its derivatives exist and are `0` when `x = +-2`. Additionally, `h(0) = e^0 = 1` ...

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

Given `f(x) = x^3-3x`

Define

`g(x) = { (f(x), " 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 (-2, 2)) :}`

The idea here is that `g(x) = f(x)` outside the interval `(-2, 2)` and in the middle of the interval `(-2, 2)`, `g(x)` 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:

`(1+x^2/(x^2-4)^2+x^4/(2(x^2-4)^4))`

is a sufficiently good approximation to `e^(x^2/(x^2-4)^2)` without itself being an exponential function.

As a result, we do not lose the smoothing properties of `h(x)` at the 'joins' at `x=+-2`

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