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)#?

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)#?

Aurora Alvarado · Accepted Answer

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

#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.

Charles Anctil · Answer

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

#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)#

Ezra Adams · Answer

See below.

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.

Emily Ammerman · Answer

#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)) :}#

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.

Christopher Agresta · Answer

undefined To construct an infinitely differentiable one-to-one function $ g(x): \mathbb{R} \rightarrow \mathbb{R} $ with $ g(x) = f(x) $ in $ (-\infty, -2] \cup [2, \infty) $, you can use the following steps:

1. Define a function $ h(x) $ as follows:
   $$
   h(x) = \begin{cases}
   f(x) & \text{if } x \leq -2 \
   f(-x) & \text{if } x \geq 2 \
   \end{cases}
   $$

2. Next, ensure that $ h(x) $ is differentiable in its domain $ (-\infty, -2] \cup [2, \infty) $. The function $ h(x) $ is continuous everywhere since it's made up of continuous pieces.

3. Now, let's construct a smooth connection between the pieces at $ x = -2 $ and $ x = 2 $. Define a function $ k(x) $ such that:
   $$
   k(x) = \begin{cases}
   0 & \text{if } x \leq -2 \
   \text{smooth transition} & \text{if } -2 < x < 2 \
   0 & \text{if } x \geq 2 \
   \end{cases}
   $$

4. The function $ k(x) $ should smoothly transition from 0 to 1 as $ x $ moves from $ -2 $ to $ 2 $. This smooth transition ensures that the function is differentiable.

5. Finally, define $ g(x) $ as the sum of $ h(x) $ and $ k(x) $:
   $$
   g(x) = h(x) + k(x)
   $$

6. This function $ g(x) $ will be infinitely differentiable and one-to-one on $ \mathbb{R} $.

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

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)#?